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Tangent to a Circle Questions in English
Class 10 Mathematics · Circles · Tangent to a Circle
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1
Easy
Fill in the blanks:
$(i)$ $A$ tangent to a circle intersects it in $.........$ point$(s)$.
$(ii)$ $A$ line intersecting a circle in two points is called a $.........$
$(i)$ $A$ tangent to a circle intersects it in $.........$ point$(s)$.
$(ii)$ $A$ line intersecting a circle in two points is called a $.........$
Solution
(A) $(i)$ $A$ tangent to a circle touches the circle at exactly one point. Therefore,it intersects the circle in $1$ point.
$(ii)$ $A$ line that intersects a circle at two distinct points is known as a secant of the circle.
$(ii)$ $A$ line that intersects a circle at two distinct points is known as a secant of the circle.
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View Solution2
Medium
Fill in the blanks:
$(i)$ $A$ circle can have $......$ parallel tangents at the most.
$(ii)$ The common point of a tangent to a circle and the circle is called $......$
$(i)$ $A$ circle can have $......$ parallel tangents at the most.
$(ii)$ The common point of a tangent to a circle and the circle is called $......$
Solution
(A-D) $(i)$ $A$ circle can have at most $2$ parallel tangents. These tangents are parallel to each other and pass through the endpoints of a diameter.
$(ii)$ The common point of a tangent to a circle and the circle is called the point of contact.
$(ii)$ The common point of a tangent to a circle and the circle is called the point of contact.
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View Solution3
MediumMCQ
$A$ tangent $PQ$ at a point $P$ of a circle of radius $5 \, cm$ meets a line through the centre $O$ at a point $Q$ so that $OQ = 12 \, cm$. The length of $PQ$ is (in $cm$):
A
$12$
B
$13$
C
$8.5$
D
$\sqrt{119}$
Solution
(D) We know that the radius of a circle is perpendicular to the tangent at the point of contact.
Therefore,$OP \perp PQ$.
This forms a right-angled triangle $\triangle OPQ$,where $OP = 5 \, cm$ (radius) and $OQ = 12 \, cm$ (hypotenuse).
By applying the Pythagoras theorem in $\triangle OPQ$:
$OP^2 + PQ^2 = OQ^2$
$5^2 + PQ^2 = 12^2$
$25 + PQ^2 = 144$
$PQ^2 = 144 - 25$
$PQ^2 = 119$
$PQ = \sqrt{119} \, cm$.
Therefore,$OP \perp PQ$.
This forms a right-angled triangle $\triangle OPQ$,where $OP = 5 \, cm$ (radius) and $OQ = 12 \, cm$ (hypotenuse).
By applying the Pythagoras theorem in $\triangle OPQ$:
$OP^2 + PQ^2 = OQ^2$
$5^2 + PQ^2 = 12^2$
$25 + PQ^2 = 144$
$PQ^2 = 144 - 25$
$PQ^2 = 119$
$PQ = \sqrt{119} \, cm$.
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View Solution4
Medium
Draw a circle and two lines parallel to a given line such that one is a tangent and the other is a secant to the circle.
Solution
(N/A) It can be observed from the figure that $AB$ and $CD$ are two parallel lines.
Line $AB$ intersects the circle at exactly two points,$P$ and $Q$. Therefore,line $AB$ is the secant of this circle.
Since line $CD$ intersects the circle at exactly one point,$R$,line $CD$ is the tangent to the circle.
Line $AB$ intersects the circle at exactly two points,$P$ and $Q$. Therefore,line $AB$ is the secant of this circle.
Since line $CD$ intersects the circle at exactly one point,$R$,line $CD$ is the tangent to the circle.
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View Solution5
Easy
Prove that in two concentric circles,the chord of the larger circle,which touches the smaller circle,is bisected at the point of contact.
Solution
(N/A) We are given two concentric circles $C_1$ and $C_2$ with centre $O$ and a chord $AB$ of the larger circle $C_1$ which touches the smaller circle $C_2$ at the point $P$. We need to prove that $AP = BP$.
Let us join $OP$. Then,$AB$ is a tangent to $C_2$ at $P$ and $OP$ is its radius.
Since the tangent at any point of a circle is perpendicular to the radius through the point of contact,we have:
$OP \perp AB$
Now,$AB$ is a chord of the circle $C_1$ and $OP \perp AB$. We know that the perpendicular drawn from the centre of a circle to a chord bisects the chord.
Therefore,$OP$ is the bisector of the chord $AB$,which implies:
$AP = BP$
Let us join $OP$. Then,$AB$ is a tangent to $C_2$ at $P$ and $OP$ is its radius.
Since the tangent at any point of a circle is perpendicular to the radius through the point of contact,we have:
$OP \perp AB$
Now,$AB$ is a chord of the circle $C_1$ and $OP \perp AB$. We know that the perpendicular drawn from the centre of a circle to a chord bisects the chord.
Therefore,$OP$ is the bisector of the chord $AB$,which implies:
$AP = BP$
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View Solution6
Difficult
Two tangents $TP$ and $TQ$ are drawn to a circle with centre $O$ from an external point $T$. Prove that $\angle PTQ = 2 \angle OPQ$.
Solution
(N/A) We are given a circle with centre $O$,an external point $T$,and two tangents $TP$ and $TQ$ to the circle,where $P$ and $Q$ are the points of contact (see figure).
We need to prove that $\angle PTQ = 2 \angle OPQ$.
Let $\angle PTQ = \theta$.
Now,$TP = TQ$ (tangents from an external point are equal in length). So,$\triangle TPQ$ is an isosceles triangle.
Therefore,$\angle TPQ = \angle TQP = \frac{1}{2} (180^{\circ} - \theta) = 90^{\circ} - \frac{1}{2} \theta$.
Also,the radius is perpendicular to the tangent at the point of contact,so $\angle OPT = 90^{\circ}$.
Thus,$\angle OPQ = \angle OPT - \angle TPQ = 90^{\circ} - (90^{\circ} - \frac{1}{2} \theta)$.
$\angle OPQ = \frac{1}{2} \theta = \frac{1}{2} \angle PTQ$.
This gives $\angle PTQ = 2 \angle OPQ$.
We need to prove that $\angle PTQ = 2 \angle OPQ$.
Let $\angle PTQ = \theta$.
Now,$TP = TQ$ (tangents from an external point are equal in length). So,$\triangle TPQ$ is an isosceles triangle.
Therefore,$\angle TPQ = \angle TQP = \frac{1}{2} (180^{\circ} - \theta) = 90^{\circ} - \frac{1}{2} \theta$.
Also,the radius is perpendicular to the tangent at the point of contact,so $\angle OPT = 90^{\circ}$.
Thus,$\angle OPQ = \angle OPT - \angle TPQ = 90^{\circ} - (90^{\circ} - \frac{1}{2} \theta)$.
$\angle OPQ = \frac{1}{2} \theta = \frac{1}{2} \angle PTQ$.
This gives $\angle PTQ = 2 \angle OPQ$.
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View Solution7
Difficult
$PQ$ is a chord of length $8 \, cm$ of a circle of radius $5 \, cm$. The tangents at $P$ and $Q$ intersect at a point $T$ (see figure). Find the length $TP$.
Solution
(D) Join $OT$. Let it intersect $PQ$ at the point $R$. Then $\triangle TPQ$ is isosceles and $TO$ is the angle bisector of $\angle PTQ$. So,$OT \perp PQ$ and therefore,$OT$ bisects $PQ$ which gives $PR = RQ = 4 \, cm$.
Also,$OR = \sqrt{OP^2 - PR^2} = \sqrt{5^2 - 4^2} \, cm = \sqrt{25 - 16} \, cm = \sqrt{9} \, cm = 3 \, cm$.
Now,$\angle TPR + \angle RPO = 90^{\circ}$ and $\angle TPR + \angle PTR = 90^{\circ}$ (since $\triangle TRP$ is a right triangle).
So,$\angle RPO = \angle PTR$.
Therefore,right triangle $TRP$ is similar to the right triangle $PRO$ by $AA$ similarity.
This gives $\frac{TP}{PO} = \frac{RP}{RO}$,i.e.,$\frac{TP}{5} = \frac{4}{3}$ or $TP = \frac{20}{3} \, cm$.
Also,$OR = \sqrt{OP^2 - PR^2} = \sqrt{5^2 - 4^2} \, cm = \sqrt{25 - 16} \, cm = \sqrt{9} \, cm = 3 \, cm$.
Now,$\angle TPR + \angle RPO = 90^{\circ}$ and $\angle TPR + \angle PTR = 90^{\circ}$ (since $\triangle TRP$ is a right triangle).
So,$\angle RPO = \angle PTR$.
Therefore,right triangle $TRP$ is similar to the right triangle $PRO$ by $AA$ similarity.
This gives $\frac{TP}{PO} = \frac{RP}{RO}$,i.e.,$\frac{TP}{5} = \frac{4}{3}$ or $TP = \frac{20}{3} \, cm$.
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View Solution8
DifficultMCQ
From a point $Q$,the length of the tangent to a circle is $24 \, cm$ and the distance of $Q$ from the centre is $25 \, cm$. The radius of the circle is (in $cm$)
A
$7$
B
$12$
C
$15$
D
$24.5$
Solution
(A) Let $O$ be the centre of the circle.
Given that,
$OQ = 25 \, cm$ and $PQ = 24 \, cm$.
As the radius is perpendicular to the tangent at the point of contact,therefore $OP \perp PQ$.
Applying Pythagoras theorem in $\triangle OPQ$,we obtain:
$OP^2 + PQ^2 = OQ^2$
$OP^2 + 24^2 = 25^2$
$OP^2 + 576 = 625$
$OP^2 = 625 - 576$
$OP^2 = 49$
$OP = 7 \, cm$.
Therefore,the radius of the circle is $7 \, cm$.
Given that,
$OQ = 25 \, cm$ and $PQ = 24 \, cm$.
As the radius is perpendicular to the tangent at the point of contact,therefore $OP \perp PQ$.
Applying Pythagoras theorem in $\triangle OPQ$,we obtain:
$OP^2 + PQ^2 = OQ^2$
$OP^2 + 24^2 = 25^2$
$OP^2 + 576 = 625$
$OP^2 = 625 - 576$
$OP^2 = 49$
$OP = 7 \, cm$.
Therefore,the radius of the circle is $7 \, cm$.
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View Solution9
DifficultMCQ
In the figure,if $TP$ and $TQ$ are the two tangents to a circle with centre $O$ such that $\angle POQ = 110^{\circ}$,then $\angle PTQ$ is equal to: (in $^{\circ}$)
A
$60$
B
$70$
C
$80$
D
$90$
Solution
(B) It is given that $TP$ and $TQ$ are tangents to the circle from an external point $T$.
The radius of a circle is perpendicular to the tangent at the point of contact.
Therefore,$OP \perp TP$ and $OQ \perp TQ$.
This implies $\angle OPT = 90^{\circ}$ and $\angle OQT = 90^{\circ}$.
In the quadrilateral $POQT$,the sum of all interior angles is $360^{\circ}$.
$\angle OPT + \angle POQ + \angle OQT + \angle PTQ = 360^{\circ}$
Substituting the known values:
$90^{\circ} + 110^{\circ} + 90^{\circ} + \angle PTQ = 360^{\circ}$
$290^{\circ} + \angle PTQ = 360^{\circ}$
$\angle PTQ = 360^{\circ} - 290^{\circ} = 70^{\circ}$
The radius of a circle is perpendicular to the tangent at the point of contact.
Therefore,$OP \perp TP$ and $OQ \perp TQ$.
This implies $\angle OPT = 90^{\circ}$ and $\angle OQT = 90^{\circ}$.
In the quadrilateral $POQT$,the sum of all interior angles is $360^{\circ}$.
$\angle OPT + \angle POQ + \angle OQT + \angle PTQ = 360^{\circ}$
Substituting the known values:
$90^{\circ} + 110^{\circ} + 90^{\circ} + \angle PTQ = 360^{\circ}$
$290^{\circ} + \angle PTQ = 360^{\circ}$
$\angle PTQ = 360^{\circ} - 290^{\circ} = 70^{\circ}$
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View Solution10
Medium
Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
Solution
(N/A) Let $AB$ be a diameter of the circle with center $O$. Two tangents $PQ$ and $RS$ are drawn at points $A$ and $B$ respectively.
Since the radius is perpendicular to the tangent at the point of contact,we have:
$OA \perp RS$ and $OB \perp PQ$
Therefore:
$\angle OAR = 90^{\circ}$
$\angle OAS = 90^{\circ}$
$\angle OBP = 90^{\circ}$
$\angle OBQ = 90^{\circ}$
From the above,we can observe that:
$\angle OAR = \angle OBQ = 90^{\circ}$ (These are alternate interior angles)
$\angle OAS = \angle OBP = 90^{\circ}$ (These are alternate interior angles)
Since the alternate interior angles are equal,the lines $PQ$ and $RS$ must be parallel.
Since the radius is perpendicular to the tangent at the point of contact,we have:
$OA \perp RS$ and $OB \perp PQ$
Therefore:
$\angle OAR = 90^{\circ}$
$\angle OAS = 90^{\circ}$
$\angle OBP = 90^{\circ}$
$\angle OBQ = 90^{\circ}$
From the above,we can observe that:
$\angle OAR = \angle OBQ = 90^{\circ}$ (These are alternate interior angles)
$\angle OAS = \angle OBP = 90^{\circ}$ (These are alternate interior angles)
Since the alternate interior angles are equal,the lines $PQ$ and $RS$ must be parallel.
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View Solution11
Difficult
Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.
Solution
(N/A) Let us consider a circle with centre $O$. Let $AB$ be a tangent which touches the circle at $P$.
We have to prove that the line perpendicular to $AB$ at $P$ passes through the centre $O$. We shall prove this by the contradiction method.
Let us assume that the perpendicular to $AB$ at $P$ does not pass through the centre $O$. Let it pass through another point $O'$.
Join $OP$ and $O'P$.
As the perpendicular to $AB$ at $P$ passes through $O'$,therefore,
$\angle O'PB = 90^{\circ} \dots(1)$
$O$ is the centre of the circle and $P$ is the point of contact. We know that the radius through the point of contact is perpendicular to the tangent.
$\therefore \angle OPB = 90^{\circ} \dots(2)$
Comparing equations $(1)$ and $(2)$,we obtain
$\angle O'PB = \angle OPB \dots(3)$
From the figure,it can be observed that,
$\angle O'PB < \angle OPB \dots(4)$
Therefore,$\angle O'PB = \angle OPB$ is not possible.
It is only possible when the line $O'P$ coincides with $OP$.
Therefore,the perpendicular to $AB$ through $P$ passes through the centre $O$.
We have to prove that the line perpendicular to $AB$ at $P$ passes through the centre $O$. We shall prove this by the contradiction method.
Let us assume that the perpendicular to $AB$ at $P$ does not pass through the centre $O$. Let it pass through another point $O'$.
Join $OP$ and $O'P$.
As the perpendicular to $AB$ at $P$ passes through $O'$,therefore,
$\angle O'PB = 90^{\circ} \dots(1)$
$O$ is the centre of the circle and $P$ is the point of contact. We know that the radius through the point of contact is perpendicular to the tangent.
$\therefore \angle OPB = 90^{\circ} \dots(2)$
Comparing equations $(1)$ and $(2)$,we obtain
$\angle O'PB = \angle OPB \dots(3)$
From the figure,it can be observed that,
$\angle O'PB < \angle OPB \dots(4)$
Therefore,$\angle O'PB = \angle OPB$ is not possible.
It is only possible when the line $O'P$ coincides with $OP$.
Therefore,the perpendicular to $AB$ through $P$ passes through the centre $O$.
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View Solution12
DifficultMCQ
The length of a tangent from a point $A$ at a distance of $5 \, cm$ from the centre of the circle is $4 \, cm$. Find the radius of the circle. (in $, cm$)
A
$16$
B
$3$
C
$9$
D
$25$
Solution
(B) Let the centre of the circle be $O$ and the point from which the tangent is drawn be $A$.
Let $B$ be the point of contact of the tangent on the circle.
Given that the distance of point $A$ from the centre $O$ is $OA = 5 \, cm$.
The length of the tangent $AB = 4 \, cm$.
We know that the radius of a circle is perpendicular to the tangent at the point of contact.
Therefore,$OB \perp AB$,which makes $\triangle ABO$ a right-angled triangle with $\angle OBA = 90^\circ$.
Applying the Pythagoras theorem in $\triangle ABO$:
$OA^2 = AB^2 + OB^2$
$5^2 = 4^2 + OB^2$
$25 = 16 + OB^2$
$OB^2 = 25 - 16$
$OB^2 = 9$
$OB = \sqrt{9} = 3 \, cm$.
Thus,the radius of the circle is $3 \, cm$.
Let $B$ be the point of contact of the tangent on the circle.
Given that the distance of point $A$ from the centre $O$ is $OA = 5 \, cm$.
The length of the tangent $AB = 4 \, cm$.
We know that the radius of a circle is perpendicular to the tangent at the point of contact.
Therefore,$OB \perp AB$,which makes $\triangle ABO$ a right-angled triangle with $\angle OBA = 90^\circ$.
Applying the Pythagoras theorem in $\triangle ABO$:
$OA^2 = AB^2 + OB^2$
$5^2 = 4^2 + OB^2$
$25 = 16 + OB^2$
$OB^2 = 25 - 16$
$OB^2 = 9$
$OB = \sqrt{9} = 3 \, cm$.
Thus,the radius of the circle is $3 \, cm$.
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View Solution13
MediumMCQ
Two concentric circles are of radii $5 \, cm$ and $3 \, cm$. Find the length of the chord of the larger circle which touches the smaller circle.
A
$16$
B
$4$
C
$8$
D
$64$
Solution
(C) Let the two concentric circles be centered at point $O$. Let $PQ$ be the chord of the larger circle which touches the smaller circle at point $A$. Therefore,$PQ$ is tangent to the smaller circle at point $A$.
Since $OA$ is the radius of the smaller circle and $PQ$ is the tangent,$OA \perp PQ$.
Applying the Pythagoras theorem in $\triangle OAP$,we have:
$OA^2 + AP^2 = OP^2$
$3^2 + AP^2 = 5^2$
$9 + AP^2 = 25$
$AP^2 = 25 - 9 = 16$
$AP = \sqrt{16} = 4 \, cm$.
Since the perpendicular from the center of a circle to a chord bisects the chord,we have $AP = AQ$.
Therefore,the length of the chord $PQ = AP + AQ = 2 \times AP = 2 \times 4 = 8 \, cm$.
Since $OA$ is the radius of the smaller circle and $PQ$ is the tangent,$OA \perp PQ$.
Applying the Pythagoras theorem in $\triangle OAP$,we have:
$OA^2 + AP^2 = OP^2$
$3^2 + AP^2 = 5^2$
$9 + AP^2 = 25$
$AP^2 = 25 - 9 = 16$
$AP = \sqrt{16} = 4 \, cm$.
Since the perpendicular from the center of a circle to a chord bisects the chord,we have $AP = AQ$.
Therefore,the length of the chord $PQ = AP + AQ = 2 \times AP = 2 \times 4 = 8 \, cm$.
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View Solution14
Easy
$A$ quadrilateral $ABCD$ is drawn to circumscribe a circle (see figure). Prove that $AB + CD = AD + BC$.
Solution
(N/A) We know that the lengths of tangents drawn from an external point to a circle are equal.
From point $D$,$DR = DS$ ... $(1)$
From point $C$,$CR = CQ$ ... $(2)$
From point $B$,$BP = BQ$ ... $(3)$
From point $A$,$AP = AS$ ... $(4)$
Adding equations $(1), (2), (3),$ and $(4)$,we get:
$DR + CR + BP + AP = DS + CQ + BQ + AS$
Rearranging the terms,we get:
$(DR + CR) + (BP + AP) = (DS + AS) + (CQ + BQ)$
Since $DR + CR = CD$,$BP + AP = AB$,$DS + AS = AD$,and $CQ + BQ = BC$,we have:
$CD + AB = AD + BC$
Hence,$AB + CD = AD + BC$.
From point $D$,$DR = DS$ ... $(1)$
From point $C$,$CR = CQ$ ... $(2)$
From point $B$,$BP = BQ$ ... $(3)$
From point $A$,$AP = AS$ ... $(4)$
Adding equations $(1), (2), (3),$ and $(4)$,we get:
$DR + CR + BP + AP = DS + CQ + BQ + AS$
Rearranging the terms,we get:
$(DR + CR) + (BP + AP) = (DS + AS) + (CQ + BQ)$
Since $DR + CR = CD$,$BP + AP = AB$,$DS + AS = AD$,and $CQ + BQ = BC$,we have:
$CD + AB = AD + BC$
Hence,$AB + CD = AD + BC$.
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View Solution15
Medium
In $Fig.$,$XY$ and $X'Y'$ are two parallel tangents to a circle with centre $O$ and another tangent $AB$ with point of contact $C$ intersecting $XY$ at $A$ and $X'Y'$ at $B$. Prove that $\angle AOB = 90^{\circ}$.
Solution
(N/A) Join point $O$ to $C$.
In $\triangle OPA$ and $\triangle OCA$:
$OP = OC$ (Radii of the same circle)
$AP = AC$ (Tangents drawn from an external point $A$ are equal in length)
$AO = AO$ (Common side)
Therefore,$\triangle OPA \cong \triangle OCA$ by $SSS$ congruence criterion.
This implies $\angle POA = \angle COA$ ... $(i)$
Similarly,in $\triangle OQB$ and $\triangle OCB$:
$OQ = OC$ (Radii of the same circle)
$BQ = BC$ (Tangents drawn from an external point $B$ are equal in length)
$OB = OB$ (Common side)
Therefore,$\triangle OQB \cong \triangle OCB$ by $SSS$ congruence criterion.
This implies $\angle QOB = \angle COB$ ... $(ii)$
Since $POQ$ is a diameter of the circle,it is a straight line,so the sum of angles on one side is $180^{\circ}$:
$\angle POA + \angle COA + \angle COB + \angle QOB = 180^{\circ}$
Using equations $(i)$ and $(ii)$:
$2 \angle COA + 2 \angle COB = 180^{\circ}$
$2(\angle COA + \angle COB) = 180^{\circ}$
$\angle COA + \angle COB = 90^{\circ}$
Therefore,$\angle AOB = 90^{\circ}$.
In $\triangle OPA$ and $\triangle OCA$:
$OP = OC$ (Radii of the same circle)
$AP = AC$ (Tangents drawn from an external point $A$ are equal in length)
$AO = AO$ (Common side)
Therefore,$\triangle OPA \cong \triangle OCA$ by $SSS$ congruence criterion.
This implies $\angle POA = \angle COA$ ... $(i)$
Similarly,in $\triangle OQB$ and $\triangle OCB$:
$OQ = OC$ (Radii of the same circle)
$BQ = BC$ (Tangents drawn from an external point $B$ are equal in length)
$OB = OB$ (Common side)
Therefore,$\triangle OQB \cong \triangle OCB$ by $SSS$ congruence criterion.
This implies $\angle QOB = \angle COB$ ... $(ii)$
Since $POQ$ is a diameter of the circle,it is a straight line,so the sum of angles on one side is $180^{\circ}$:
$\angle POA + \angle COA + \angle COB + \angle QOB = 180^{\circ}$
Using equations $(i)$ and $(ii)$:
$2 \angle COA + 2 \angle COB = 180^{\circ}$
$2(\angle COA + \angle COB) = 180^{\circ}$
$\angle COA + \angle COB = 90^{\circ}$
Therefore,$\angle AOB = 90^{\circ}$.
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View Solution16
Difficult
Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segment joining the points of contact at the centre.
Solution
(N/A) Let us consider a circle centered at point $O$. Let $P$ be an external point from which two tangents $PA$ and $PB$ are drawn to the circle,touching the circle at points $A$ and $B$ respectively. $AB$ is the line segment joining the points of contact $A$ and $B$,which subtends $\angle AOB$ at the center $O$ of the circle.
It can be observed that:
$OA$ (radius) $\perp PA$ (tangent)
Therefore,$\angle OAP = 90^{\circ}$
Similarly,$OB$ (radius) $\perp PB$ (tangent)
Therefore,$\angle OBP = 90^{\circ}$
In the quadrilateral $OAPB$,the sum of all interior angles is $360^{\circ}$.
$\angle OAP + \angle APB + \angle PBO + \angle BOA = 360^{\circ}$
$90^{\circ} + \angle APB + 90^{\circ} + \angle BOA = 360^{\circ}$
$\angle APB + \angle BOA = 180^{\circ}$
Hence,it is proved that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segment joining the points of contact at the center.
It can be observed that:
$OA$ (radius) $\perp PA$ (tangent)
Therefore,$\angle OAP = 90^{\circ}$
Similarly,$OB$ (radius) $\perp PB$ (tangent)
Therefore,$\angle OBP = 90^{\circ}$
In the quadrilateral $OAPB$,the sum of all interior angles is $360^{\circ}$.
$\angle OAP + \angle APB + \angle PBO + \angle BOA = 360^{\circ}$
$90^{\circ} + \angle APB + 90^{\circ} + \angle BOA = 360^{\circ}$
$\angle APB + \angle BOA = 180^{\circ}$
Hence,it is proved that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segment joining the points of contact at the center.
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View Solution17
Difficult
Prove that the parallelogram circumscribing a circle is a rhombus.
Solution
(N/A) Since $ABCD$ is a parallelogram,
$AB = CD \dots(1)$
$BC = AD \dots(2)$
It can be observed that:
$DR = DS$ (Tangents on the circle from point $D$)
$CR = CQ$ (Tangents on the circle from point $C$)
$BP = BQ$ (Tangents on the circle from point $B$)
$AP = AS$ (Tangents on the circle from point $A$)
Adding all these equations,we obtain:
$DR + CR + BP + AP = DS + CQ + BQ + AS$
$(DR + CR) + (BP + AP) = (DS + AS) + (CQ + BQ)$
$CD + AB = AD + BC$
On putting the values of equations $(1)$ and $(2)$ in this equation,we obtain:
$2AB = 2BC$
$AB = BC \dots(3)$
Comparing equations $(1), (2),$ and $(3),$ we obtain:
$AB = BC = CD = DA$
Hence,$ABCD$ is a rhombus.
$AB = CD \dots(1)$
$BC = AD \dots(2)$
It can be observed that:
$DR = DS$ (Tangents on the circle from point $D$)
$CR = CQ$ (Tangents on the circle from point $C$)
$BP = BQ$ (Tangents on the circle from point $B$)
$AP = AS$ (Tangents on the circle from point $A$)
Adding all these equations,we obtain:
$DR + CR + BP + AP = DS + CQ + BQ + AS$
$(DR + CR) + (BP + AP) = (DS + AS) + (CQ + BQ)$
$CD + AB = AD + BC$
On putting the values of equations $(1)$ and $(2)$ in this equation,we obtain:
$2AB = 2BC$
$AB = BC \dots(3)$
Comparing equations $(1), (2),$ and $(3),$ we obtain:
$AB = BC = CD = DA$
Hence,$ABCD$ is a rhombus.
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View Solution18
DifficultMCQ
$A$ triangle $ABC$ is drawn to circumscribe a circle of radius $4 \, cm$ such that the segments $BD$ and $DC$ into which $BC$ is divided by the point of contact $D$ are of lengths $8 \, cm$ and $6 \, cm$ respectively (see figure). Find the sides $AB$ and $AC$ (in $cm$).
A
$7, 14$
B
$20, 22$
C
$95, 97$
D
$15, 13$
Solution
(D) Let the circle touch the sides $AB$ and $AC$ of the triangle at points $E$ and $F$ respectively. Let the length of the tangent $AF$ be $x \, cm$.
Since tangents drawn from an external point to a circle are equal in length:
$CF = CD = 6 \, cm$
$BE = BD = 8 \, cm$
$AE = AF = x \, cm$
Thus,the sides of the triangle are:
$AB = AE + EB = x + 8$
$BC = BD + DC = 8 + 6 = 14 \, cm$
$AC = AF + FC = x + 6$
Semi-perimeter $s = \frac{(x+8) + 14 + (x+6)}{2} = \frac{2x + 28}{2} = x + 14$.
Using Heron's formula,Area of $\triangle ABC = \sqrt{s(s-a)(s-b)(s-c)}$
$= \sqrt{(x+14)(x+14-14)(x+14-(x+6))(x+14-(x+8))}$
$= \sqrt{(x+14)(x)(8)(6)} = \sqrt{48x(x+14)} = 4\sqrt{3x(x+14)}$.
Also,Area of $\triangle ABC = \text{Area}(\triangle OBC) + \text{Area}(\triangle OCA) + \text{Area}(\triangle OAB)$
$= \frac{1}{2} \times 4 \times 14 + \frac{1}{2} \times 4 \times (x+6) + \frac{1}{2} \times 4 \times (x+8)$
$= 28 + 2(x+6) + 2(x+8) = 28 + 2x + 12 + 2x + 16 = 56 + 4x = 4(14+x)$.
Equating the areas: $4\sqrt{3x(x+14)} = 4(14+x)$
$\sqrt{3x(x+14)} = 14+x$
Squaring both sides: $3x(x+14) = (14+x)^2$
$3x^2 + 42x = 196 + x^2 + 28x$
$2x^2 + 14x - 196 = 0 \Rightarrow x^2 + 7x - 98 = 0$
$(x+14)(x-7) = 0$. Since $x > 0$,$x = 7$.
Therefore,$AB = 7+8 = 15 \, cm$ and $AC = 7+6 = 13 \, cm$.
Since tangents drawn from an external point to a circle are equal in length:
$CF = CD = 6 \, cm$
$BE = BD = 8 \, cm$
$AE = AF = x \, cm$
Thus,the sides of the triangle are:
$AB = AE + EB = x + 8$
$BC = BD + DC = 8 + 6 = 14 \, cm$
$AC = AF + FC = x + 6$
Semi-perimeter $s = \frac{(x+8) + 14 + (x+6)}{2} = \frac{2x + 28}{2} = x + 14$.
Using Heron's formula,Area of $\triangle ABC = \sqrt{s(s-a)(s-b)(s-c)}$
$= \sqrt{(x+14)(x+14-14)(x+14-(x+6))(x+14-(x+8))}$
$= \sqrt{(x+14)(x)(8)(6)} = \sqrt{48x(x+14)} = 4\sqrt{3x(x+14)}$.
Also,Area of $\triangle ABC = \text{Area}(\triangle OBC) + \text{Area}(\triangle OCA) + \text{Area}(\triangle OAB)$
$= \frac{1}{2} \times 4 \times 14 + \frac{1}{2} \times 4 \times (x+6) + \frac{1}{2} \times 4 \times (x+8)$
$= 28 + 2(x+6) + 2(x+8) = 28 + 2x + 12 + 2x + 16 = 56 + 4x = 4(14+x)$.
Equating the areas: $4\sqrt{3x(x+14)} = 4(14+x)$
$\sqrt{3x(x+14)} = 14+x$
Squaring both sides: $3x(x+14) = (14+x)^2$
$3x^2 + 42x = 196 + x^2 + 28x$
$2x^2 + 14x - 196 = 0 \Rightarrow x^2 + 7x - 98 = 0$
$(x+14)(x-7) = 0$. Since $x > 0$,$x = 7$.
Therefore,$AB = 7+8 = 15 \, cm$ and $AC = 7+6 = 13 \, cm$.
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View Solution19
Difficult
Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
Solution
(N/A) Let $ABCD$ be a quadrilateral circumscribing a circle centered at $O$ such that it touches the circle at points $P, Q, R, S$. Let us join the vertices of the quadrilateral $ABCD$ to the center of the circle.
Consider $\triangle OAP$ and $\triangle OAS$:
$AP = AS$ (Tangents from the same point)
$OP = OS$ (Radii of the same circle)
$OA = OA$ (Common side)
Therefore,$\triangle OAP \cong \triangle OAS$ ($SSS$ congruence criterion).
Thus,$\angle POA = \angle AOS$,which implies $\angle 1 = \angle 8$.
Similarly,we can show that:
$\angle 2 = \angle 3$
$\angle 4 = \angle 5$
$\angle 6 = \angle 7$
Since the sum of angles around the center is $360^{\circ}$:
$\angle 1 + \angle 2 + \angle 3 + \angle 4 + \angle 5 + \angle 6 + \angle 7 + \angle 8 = 360^{\circ}$
Substituting the equal angles:
$(\angle 1 + \angle 8) + (\angle 2 + \angle 3) + (\angle 4 + \angle 5) + (\angle 6 + \angle 7) = 360^{\circ}$
$2\angle 1 + 2\angle 2 + 2\angle 5 + 2\angle 6 = 360^{\circ}$
$2(\angle 1 + \angle 2) + 2(\angle 5 + \angle 6) = 360^{\circ}$
$(\angle 1 + \angle 2) + (\angle 5 + \angle 6) = 180^{\circ}$
Thus,$\angle AOB + \angle COD = 180^{\circ}$.
Similarly,we can prove that $\angle BOC + \angle DOA = 180^{\circ}$.
Hence,opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the center of the circle.
Consider $\triangle OAP$ and $\triangle OAS$:
$AP = AS$ (Tangents from the same point)
$OP = OS$ (Radii of the same circle)
$OA = OA$ (Common side)
Therefore,$\triangle OAP \cong \triangle OAS$ ($SSS$ congruence criterion).
Thus,$\angle POA = \angle AOS$,which implies $\angle 1 = \angle 8$.
Similarly,we can show that:
$\angle 2 = \angle 3$
$\angle 4 = \angle 5$
$\angle 6 = \angle 7$
Since the sum of angles around the center is $360^{\circ}$:
$\angle 1 + \angle 2 + \angle 3 + \angle 4 + \angle 5 + \angle 6 + \angle 7 + \angle 8 = 360^{\circ}$
Substituting the equal angles:
$(\angle 1 + \angle 8) + (\angle 2 + \angle 3) + (\angle 4 + \angle 5) + (\angle 6 + \angle 7) = 360^{\circ}$
$2\angle 1 + 2\angle 2 + 2\angle 5 + 2\angle 6 = 360^{\circ}$
$2(\angle 1 + \angle 2) + 2(\angle 5 + \angle 6) = 360^{\circ}$
$(\angle 1 + \angle 2) + (\angle 5 + \angle 6) = 180^{\circ}$
Thus,$\angle AOB + \angle COD = 180^{\circ}$.
Similarly,we can prove that $\angle BOC + \angle DOA = 180^{\circ}$.
Hence,opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the center of the circle.
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View Solution20
MediumMCQ
If the radii of two concentric circles are $4 \, cm$ and $5 \, cm$,then the length of each chord of the larger circle which is tangent to the smaller circle is (in $cm$):
A
$3$
B
$9$
C
$1$
D
$6$
Solution
(D) Let $O$ be the centre of two concentric circles $C_{1}$ and $C_{2}$,whose radii are $r_{1} = 4 \, cm$ and $r_{2} = 5 \, cm$.
Let $AC$ be a chord of the larger circle $C_{2}$,which touches the smaller circle $C_{1}$ at point $B$.
Join $OB$. Since $AC$ is a tangent to the circle $C_{1}$ at $B$,$OB \perp AC$ (The tangent at any point of a circle is perpendicular to the radius through the point of contact).
Now,in the right-angled $\triangle OBC$,by using the Pythagoras theorem:
$OC^{2} = BC^{2} + OB^{2}$
Substituting the values,we get:
$5^{2} = BC^{2} + 4^{2}$
$25 = BC^{2} + 16$
$BC^{2} = 25 - 16 = 9$
$BC = \sqrt{9} = 3 \, cm$
Since the perpendicular from the centre to a chord bisects the chord,the length of the chord $AC = 2 \times BC = 2 \times 3 = 6 \, cm$.
Let $AC$ be a chord of the larger circle $C_{2}$,which touches the smaller circle $C_{1}$ at point $B$.
Join $OB$. Since $AC$ is a tangent to the circle $C_{1}$ at $B$,$OB \perp AC$ (The tangent at any point of a circle is perpendicular to the radius through the point of contact).
Now,in the right-angled $\triangle OBC$,by using the Pythagoras theorem:
$OC^{2} = BC^{2} + OB^{2}$
Substituting the values,we get:
$5^{2} = BC^{2} + 4^{2}$
$25 = BC^{2} + 16$
$BC^{2} = 25 - 16 = 9$
$BC = \sqrt{9} = 3 \, cm$
Since the perpendicular from the centre to a chord bisects the chord,the length of the chord $AC = 2 \times BC = 2 \times 3 = 6 \, cm$.
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View Solution21
MediumMCQ
In the given figure,$AT$ is a tangent to the circle with center $O$ such that $OT = 4 \, cm$ and $\angle OTA = 30^{\circ}$. Then $AT$ is equal to (in $cm$):
A
$2 \sqrt{3}$
B
$4 \sqrt{3}$
C
$2$
D
$4$
Solution
(A) In $\triangle OAT$,we know that the tangent at any point of a circle is perpendicular to the radius through the point of contact.
Therefore,$\angle OAT = 90^{\circ}$.
In the right-angled triangle $\triangle OAT$,we have:
$\cos(\angle OTA) = \frac{\text{Adjacent side}}{\text{Hypotenuse}} = \frac{AT}{OT}$
Given that $OT = 4 \, cm$ and $\angle OTA = 30^{\circ}$,we substitute these values:
$\cos(30^{\circ}) = \frac{AT}{4}$
Since $\cos(30^{\circ}) = \frac{\sqrt{3}}{2}$,we get:
$\frac{\sqrt{3}}{2} = \frac{AT}{4}$
$AT = 4 \times \frac{\sqrt{3}}{2}$
$AT = 2 \sqrt{3} \, cm$.
Therefore,$\angle OAT = 90^{\circ}$.
In the right-angled triangle $\triangle OAT$,we have:
$\cos(\angle OTA) = \frac{\text{Adjacent side}}{\text{Hypotenuse}} = \frac{AT}{OT}$
Given that $OT = 4 \, cm$ and $\angle OTA = 30^{\circ}$,we substitute these values:
$\cos(30^{\circ}) = \frac{AT}{4}$
Since $\cos(30^{\circ}) = \frac{\sqrt{3}}{2}$,we get:
$\frac{\sqrt{3}}{2} = \frac{AT}{4}$
$AT = 4 \times \frac{\sqrt{3}}{2}$
$AT = 2 \sqrt{3} \, cm$.
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View Solution22
Medium
State whether the following statement is 'True' or 'False' and provide reasons for your answer: The length of a tangent from an external point to a circle is always greater than the radius of the circle.
Solution
(B) The statement is 'False'.
Let the radius of the circle be $r$ and the length of the tangent from an external point $P$ be $L$.
In a right-angled triangle formed by the radius, the tangent, and the line segment connecting the center to the external point, the hypotenuse is the distance from the center to the external point $(d)$.
By the Pythagorean theorem, $d^2 = r^2 + L^2$.
Since $d > r$, it follows that $L = \sqrt{d^2 - r^2}$.
Depending on the distance $d$ of the external point from the center, the length of the tangent $L$ can be greater than, equal to, or less than the radius $r$.
For example, if the external point is very close to the circle, the tangent length can be smaller than the radius.
Let the radius of the circle be $r$ and the length of the tangent from an external point $P$ be $L$.
In a right-angled triangle formed by the radius, the tangent, and the line segment connecting the center to the external point, the hypotenuse is the distance from the center to the external point $(d)$.
By the Pythagorean theorem, $d^2 = r^2 + L^2$.
Since $d > r$, it follows that $L = \sqrt{d^2 - r^2}$.
Depending on the distance $d$ of the external point from the center, the length of the tangent $L$ can be greater than, equal to, or less than the radius $r$.
For example, if the external point is very close to the circle, the tangent length can be smaller than the radius.
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View Solution23
Medium
Write 'True' or 'False' and give reasons for your answer.
The length of the tangent from an external point $P$ on a circle with centre $O$ is always less than $OP$.
The length of the tangent from an external point $P$ on a circle with centre $O$ is always less than $OP$.
Solution
(TRUE) True.
Let $PT$ be the tangent drawn from an external point $P$ to the circle at point of contact $T$. Join $OT$.
Since the radius is perpendicular to the tangent at the point of contact,we have $OT \perp PT$.
Thus,$\triangle OPT$ is a right-angled triangle with $\angle OTP = 90^{\circ}$.
In a right-angled triangle,the hypotenuse is the longest side. Here,$OP$ is the hypotenuse.
Therefore,$OP > PT$,which implies that the length of the tangent $PT$ is always less than $OP$.
Let $PT$ be the tangent drawn from an external point $P$ to the circle at point of contact $T$. Join $OT$.
Since the radius is perpendicular to the tangent at the point of contact,we have $OT \perp PT$.
Thus,$\triangle OPT$ is a right-angled triangle with $\angle OTP = 90^{\circ}$.
In a right-angled triangle,the hypotenuse is the longest side. Here,$OP$ is the hypotenuse.
Therefore,$OP > PT$,which implies that the length of the tangent $PT$ is always less than $OP$.
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View Solution24
Medium
Write 'True' or 'False' and give reasons for your answer.
The tangent to the circumcircle of an isosceles triangle $ABC$ at $A$,in which $AB = AC$,is parallel to $BC$.
The tangent to the circumcircle of an isosceles triangle $ABC$ at $A$,in which $AB = AC$,is parallel to $BC$.
Solution
(A) True.
Let $EAF$ be the tangent to the circumcircle of $\triangle ABC$ at point $A$.
We need to prove that $EAF \parallel BC$.
By the alternate segment theorem,the angle between the tangent $EAF$ and the chord $AB$ is equal to the angle subtended by the chord $AB$ in the alternate segment,which is $\angle ACB$.
Thus,$\angle EAB = \angle ACB$.
Since $\triangle ABC$ is an isosceles triangle with $AB = AC$,the angles opposite to equal sides are equal,so $\angle ABC = \angle ACB$.
From these two relations,we get $\angle EAB = \angle ABC$.
Since these are alternate interior angles formed by the line $EAF$ and $BC$ with the transversal $AB$,the equality of these angles implies that $EAF \parallel BC$.
Let $EAF$ be the tangent to the circumcircle of $\triangle ABC$ at point $A$.
We need to prove that $EAF \parallel BC$.
By the alternate segment theorem,the angle between the tangent $EAF$ and the chord $AB$ is equal to the angle subtended by the chord $AB$ in the alternate segment,which is $\angle ACB$.
Thus,$\angle EAB = \angle ACB$.
Since $\triangle ABC$ is an isosceles triangle with $AB = AC$,the angles opposite to equal sides are equal,so $\angle ABC = \angle ACB$.
From these two relations,we get $\angle EAB = \angle ABC$.
Since these are alternate interior angles formed by the line $EAF$ and $BC$ with the transversal $AB$,the equality of these angles implies that $EAF \parallel BC$.
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View Solution25
Difficult
Write 'True' or 'False' and give reasons for your answer.
If a number of circles touch a given line segment $PQ$ at a point $A$,then their centres lie on the perpendicular bisector of $PQ$.
If a number of circles touch a given line segment $PQ$ at a point $A$,then their centres lie on the perpendicular bisector of $PQ$.
Solution
(B) False.
Let $PQ$ be a line segment and $A$ be a point on it. If several circles touch the line segment $PQ$ at point $A$,then the radius of each circle at the point of contact $A$ is perpendicular to the line segment $PQ$.
Let the centres of these circles be $C_1, C_2, C_3, \dots$. Since each circle touches the line segment $PQ$ at point $A$,the line segment connecting the centre of each circle to the point $A$ (i.e.,$C_1A, C_2A, C_3A, \dots$) must be perpendicular to $PQ$ at point $A$.
This means that all the centres $C_1, C_2, C_3, \dots$ lie on a line that is perpendicular to $PQ$ at point $A$. However,for these centres to lie on the perpendicular bisector of $PQ$,the point $A$ must be the midpoint of $PQ$. Since the problem statement does not specify that $A$ is the midpoint of $PQ$,the centres lie on a line perpendicular to $PQ$ at $A$,but not necessarily on the perpendicular bisector of $PQ$.
Let $PQ$ be a line segment and $A$ be a point on it. If several circles touch the line segment $PQ$ at point $A$,then the radius of each circle at the point of contact $A$ is perpendicular to the line segment $PQ$.
Let the centres of these circles be $C_1, C_2, C_3, \dots$. Since each circle touches the line segment $PQ$ at point $A$,the line segment connecting the centre of each circle to the point $A$ (i.e.,$C_1A, C_2A, C_3A, \dots$) must be perpendicular to $PQ$ at point $A$.
This means that all the centres $C_1, C_2, C_3, \dots$ lie on a line that is perpendicular to $PQ$ at point $A$. However,for these centres to lie on the perpendicular bisector of $PQ$,the point $A$ must be the midpoint of $PQ$. Since the problem statement does not specify that $A$ is the midpoint of $PQ$,the centres lie on a line perpendicular to $PQ$ at $A$,but not necessarily on the perpendicular bisector of $PQ$.
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View Solution26
Medium
Write 'True' or 'False' and give reasons for your answer.
$AB$ is a diameter of a circle and $AC$ is its chord such that $\angle BAC = 30^{\circ}$. If the tangent at $C$ intersects $AB$ extended at $D$,then $BC = BD$.
$AB$ is a diameter of a circle and $AC$ is its chord such that $\angle BAC = 30^{\circ}$. If the tangent at $C$ intersects $AB$ extended at $D$,then $BC = BD$.
Solution
(A) True
To prove: $BC = BD$.
$1$. Join $OC$. Since $AB$ is the diameter,$\angle ACB = 90^{\circ}$ (angle in a semicircle).
$2$. In $\triangle ABC$,$\angle ABC = 180^{\circ} - (90^{\circ} + 30^{\circ}) = 60^{\circ}$.
$3$. Since $OC$ is the radius and $CD$ is the tangent,$OC \perp CD$,so $\angle OCD = 90^{\circ}$.
$4$. In $\triangle OAC$,$OA = OC$ (radii),so $\angle OCA = \angle OAC = 30^{\circ}$.
$5$. $\angle BCD = \angle OCD - \angle OCB = 90^{\circ} - (90^{\circ} - 30^{\circ}) = 30^{\circ}$.
$6$. In $\triangle BCD$,$\angle CBD = 180^{\circ} - \angle ABC = 180^{\circ} - 60^{\circ} = 120^{\circ}$ (linear pair).
$7$. In $\triangle BCD$,$\angle BDC = 180^{\circ} - (120^{\circ} + 30^{\circ}) = 30^{\circ}$.
$8$. Since $\angle BCD = \angle BDC = 30^{\circ}$,the sides opposite to these angles are equal,hence $BC = BD$.
To prove: $BC = BD$.
$1$. Join $OC$. Since $AB$ is the diameter,$\angle ACB = 90^{\circ}$ (angle in a semicircle).
$2$. In $\triangle ABC$,$\angle ABC = 180^{\circ} - (90^{\circ} + 30^{\circ}) = 60^{\circ}$.
$3$. Since $OC$ is the radius and $CD$ is the tangent,$OC \perp CD$,so $\angle OCD = 90^{\circ}$.
$4$. In $\triangle OAC$,$OA = OC$ (radii),so $\angle OCA = \angle OAC = 30^{\circ}$.
$5$. $\angle BCD = \angle OCD - \angle OCB = 90^{\circ} - (90^{\circ} - 30^{\circ}) = 30^{\circ}$.
$6$. In $\triangle BCD$,$\angle CBD = 180^{\circ} - \angle ABC = 180^{\circ} - 60^{\circ} = 120^{\circ}$ (linear pair).
$7$. In $\triangle BCD$,$\angle BDC = 180^{\circ} - (120^{\circ} + 30^{\circ}) = 30^{\circ}$.
$8$. Since $\angle BCD = \angle BDC = 30^{\circ}$,the sides opposite to these angles are equal,hence $BC = BD$.
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View Solution27
Medium
If $d_{1}$ and $d_{2}$ $(d_{2} > d_{1})$ are the diameters of two concentric circles and $c$ is the length of a chord of the outer circle which is tangent to the inner circle,prove that $d_{2}^{2} = c^{2} + d_{1}^{2}$.
Solution
(N/A) Let $O$ be the common center of the two concentric circles. Let $AB$ be a chord of the outer circle which touches the inner circle at point $C$.
Since the radius is perpendicular to the tangent at the point of contact,$OC \perp AB$.
In the right-angled triangle $\Delta OCB$,by Pythagoras theorem:
$OC^{2} + CB^{2} = OB^{2}$
Here,$OC$ is the radius of the inner circle,so $OC = \frac{d_{1}}{2}$.
$OB$ is the radius of the outer circle,so $OB = \frac{d_{2}}{2}$.
Since the perpendicular from the center to a chord bisects the chord,$CB = \frac{c}{2}$.
Substituting these values in the equation:
$(\frac{d_{1}}{2})^{2} + (\frac{c}{2})^{2} = (\frac{d_{2}}{2})^{2}$
$\frac{d_{1}^{2}}{4} + \frac{c^{2}}{4} = \frac{d_{2}^{2}}{4}$
Multiplying both sides by $4$,we get:
$d_{1}^{2} + c^{2} = d_{2}^{2}$
Hence,$d_{2}^{2} = c^{2} + d_{1}^{2}$.
Since the radius is perpendicular to the tangent at the point of contact,$OC \perp AB$.
In the right-angled triangle $\Delta OCB$,by Pythagoras theorem:
$OC^{2} + CB^{2} = OB^{2}$
Here,$OC$ is the radius of the inner circle,so $OC = \frac{d_{1}}{2}$.
$OB$ is the radius of the outer circle,so $OB = \frac{d_{2}}{2}$.
Since the perpendicular from the center to a chord bisects the chord,$CB = \frac{c}{2}$.
Substituting these values in the equation:
$(\frac{d_{1}}{2})^{2} + (\frac{c}{2})^{2} = (\frac{d_{2}}{2})^{2}$
$\frac{d_{1}^{2}}{4} + \frac{c^{2}}{4} = \frac{d_{2}^{2}}{4}$
Multiplying both sides by $4$,we get:
$d_{1}^{2} + c^{2} = d_{2}^{2}$
Hence,$d_{2}^{2} = c^{2} + d_{1}^{2}$.
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View Solution28
MediumMCQ
Out of two concentric circles,the radius of the outer circle is $5 \, cm$ and the chord $AC$ of length $8 \, cm$ is a tangent to the inner circle. Find the radius of the inner circle (in $cm$).
A
$2$
B
$1.5$
C
$9$
D
$3$
Solution
(D) Let $C_1$ and $C_2$ be the circles having the same centre $O$. $AC$ is a chord of the outer circle $C_2$ which touches the inner circle $C_1$ at point $D$.
Join $OD$.
Since $AC$ is a tangent to the inner circle at $D$,$OD \perp AC$.
$A$ perpendicular drawn from the centre to a chord bisects the chord.
Therefore,$AD = DC = \frac{8}{2} = 4 \, cm$.
In the right-angled $\triangle AOD$,by Pythagoras theorem:
$OA^2 = AD^2 + OD^2$
$5^2 = 4^2 + OD^2$
$25 = 16 + OD^2$
$OD^2 = 25 - 16 = 9$
$OD = \sqrt{9} = 3 \, cm$.
Thus,the radius of the inner circle is $3 \, cm$.
Join $OD$.
Since $AC$ is a tangent to the inner circle at $D$,$OD \perp AC$.
$A$ perpendicular drawn from the centre to a chord bisects the chord.
Therefore,$AD = DC = \frac{8}{2} = 4 \, cm$.
In the right-angled $\triangle AOD$,by Pythagoras theorem:
$OA^2 = AD^2 + OD^2$
$5^2 = 4^2 + OD^2$
$25 = 16 + OD^2$
$OD^2 = 25 - 16 = 9$
$OD = \sqrt{9} = 3 \, cm$.
Thus,the radius of the inner circle is $3 \, cm$.
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View Solution29
Difficult
$AB$ and $CD$ are common tangents to two circles. If the radii of the two circles are equal,prove that $AB = CD$.
Solution
(N/A) Given: $AB$ and $CD$ are common tangents to two circles with centers $O$ and $O^{\prime}$ and equal radii $r$.
To prove: $AB = CD$.
Construction: Join $OA, OC, O^{\prime}B$ and $O^{\prime}D$.
Proof:
$1$. Since $AB$ is a tangent to the circle at $A$,the radius $OA$ is perpendicular to $AB$. Thus,$\angle OAB = 90^{\circ}$.
$2$. Similarly,since $CD$ is a tangent to the circle at $C$,the radius $OC$ is perpendicular to $CD$. Thus,$\angle OCD = 90^{\circ}$.
$3$. Since $AB$ and $CD$ are parallel tangents (as they are perpendicular to the line joining the centers if the radii are equal),$AC$ is a diameter or a line segment perpendicular to the tangents.
$4$. In the quadrilateral $ABDC$,we have $\angle A = 90^{\circ}$,$\angle B = 90^{\circ}$,$\angle C = 90^{\circ}$,and $\angle D = 90^{\circ}$.
$5$. Since the radii are equal,the distance between the parallel tangents $AB$ and $CD$ is constant,making $AC = BD = 2r$.
$6$. $A$ quadrilateral with all angles equal to $90^{\circ}$ and opposite sides equal is a rectangle.
$7$. Therefore,$ABDC$ is a rectangle.
$8$. Hence,the opposite sides are equal,which implies $AB = CD$.
To prove: $AB = CD$.
Construction: Join $OA, OC, O^{\prime}B$ and $O^{\prime}D$.
Proof:
$1$. Since $AB$ is a tangent to the circle at $A$,the radius $OA$ is perpendicular to $AB$. Thus,$\angle OAB = 90^{\circ}$.
$2$. Similarly,since $CD$ is a tangent to the circle at $C$,the radius $OC$ is perpendicular to $CD$. Thus,$\angle OCD = 90^{\circ}$.
$3$. Since $AB$ and $CD$ are parallel tangents (as they are perpendicular to the line joining the centers if the radii are equal),$AC$ is a diameter or a line segment perpendicular to the tangents.
$4$. In the quadrilateral $ABDC$,we have $\angle A = 90^{\circ}$,$\angle B = 90^{\circ}$,$\angle C = 90^{\circ}$,and $\angle D = 90^{\circ}$.
$5$. Since the radii are equal,the distance between the parallel tangents $AB$ and $CD$ is constant,making $AC = BD = 2r$.
$6$. $A$ quadrilateral with all angles equal to $90^{\circ}$ and opposite sides equal is a rectangle.
$7$. Therefore,$ABDC$ is a rectangle.
$8$. Hence,the opposite sides are equal,which implies $AB = CD$.
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View Solution30
Medium
$A$ chord $PQ$ of a circle is parallel to the tangent drawn at a point $R$ of the circle. Prove that $R$ bisects the arc $PRQ$.
Solution
(N/A) Given: Chord $PQ$ is parallel to the tangent $MN$ at point $R$.
To prove: $R$ bisects the arc $PRQ$, i.e., arc $PR = \text{arc } RQ$.
Proof:
Let $MN$ be the tangent at $R$. Since $PQ \parallel MN$, the alternate interior angles are equal.
Therefore, $\angle MRP = \angle RPQ$ (Let this be $\angle 1 = \angle 2$).
By the alternate segment theorem, the angle between the tangent and the chord is equal to the angle subtended by the chord in the alternate segment.
Thus, $\angle MRP = \angle RQP$ (Let this be $\angle 1 = \angle 3$).
From the above two equations, we get $\angle 2 = \angle 3$, i.e., $\angle RPQ = \angle RQP$.
In $\triangle PQR$, since $\angle RPQ = \angle RQP$, the sides opposite to these angles are equal.
Therefore, $PR = RQ$.
Since the chords $PR$ and $RQ$ are equal, the arcs subtended by them are also equal.
Hence, arc $PR = \text{arc } RQ$, which means $R$ bisects the arc $PRQ$.
To prove: $R$ bisects the arc $PRQ$, i.e., arc $PR = \text{arc } RQ$.
Proof:
Let $MN$ be the tangent at $R$. Since $PQ \parallel MN$, the alternate interior angles are equal.
Therefore, $\angle MRP = \angle RPQ$ (Let this be $\angle 1 = \angle 2$).
By the alternate segment theorem, the angle between the tangent and the chord is equal to the angle subtended by the chord in the alternate segment.
Thus, $\angle MRP = \angle RQP$ (Let this be $\angle 1 = \angle 3$).
From the above two equations, we get $\angle 2 = \angle 3$, i.e., $\angle RPQ = \angle RQP$.
In $\triangle PQR$, since $\angle RPQ = \angle RQP$, the sides opposite to these angles are equal.
Therefore, $PR = RQ$.
Since the chords $PR$ and $RQ$ are equal, the arcs subtended by them are also equal.
Hence, arc $PR = \text{arc } RQ$, which means $R$ bisects the arc $PRQ$.
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View Solution31
Difficult
In a right triangle $ABC$ in which $\angle B = 90^{\circ}$,a circle is drawn with $AB$ as diameter,intersecting the hypotenuse $AC$ at $P$. Prove that the tangent to the circle at $P$ bisects $BC$.
Solution
(N/A) Let $O$ be the centre of the given circle. Suppose the tangent at $P$ meets $BC$ at $Q$. Join $BP$.
To prove: $BQ = QC$.
Proof: $\angle ABC = 90^{\circ}$.
In $\triangle ABC$,$\angle A + \angle C = 90^{\circ}$ (where $\angle A = \angle 1$ and $\angle C = \angle 5$).
So,$\angle 1 + \angle 5 = 90^{\circ}$ ... $(i)$
By the alternate segment theorem,the angle between the tangent at $P$ and the chord $BP$ is equal to the angle in the alternate segment,so $\angle 3 = \angle 1$.
Also,$\angle APB = 90^{\circ}$ (angle in a semi-circle).
Since $AC$ is a straight line,$\angle APB + \angle BPC = 180^{\circ}$,so $\angle BPC = 90^{\circ}$.
In $\triangle BPC$,$\angle 4 + \angle 5 = 90^{\circ}$ ... $(ii)$
From $(i)$ and $(ii)$:
$\angle 1 + \angle 5 = \angle 4 + \angle 5$
$\Rightarrow \angle 1 = \angle 4$
Since $\angle 3 = \angle 1$,we have $\angle 3 = \angle 4$.
In $\triangle PQC$,since $\angle 3 = \angle 4$,the sides opposite to equal angles are equal,so $PQ = QC$.
Also,tangents drawn from an external point to a circle are equal,so $PQ = BQ$.
Therefore,$BQ = QC$.
Hence,the tangent at $P$ bisects $BC$.
To prove: $BQ = QC$.
Proof: $\angle ABC = 90^{\circ}$.
In $\triangle ABC$,$\angle A + \angle C = 90^{\circ}$ (where $\angle A = \angle 1$ and $\angle C = \angle 5$).
So,$\angle 1 + \angle 5 = 90^{\circ}$ ... $(i)$
By the alternate segment theorem,the angle between the tangent at $P$ and the chord $BP$ is equal to the angle in the alternate segment,so $\angle 3 = \angle 1$.
Also,$\angle APB = 90^{\circ}$ (angle in a semi-circle).
Since $AC$ is a straight line,$\angle APB + \angle BPC = 180^{\circ}$,so $\angle BPC = 90^{\circ}$.
In $\triangle BPC$,$\angle 4 + \angle 5 = 90^{\circ}$ ... $(ii)$
From $(i)$ and $(ii)$:
$\angle 1 + \angle 5 = \angle 4 + \angle 5$
$\Rightarrow \angle 1 = \angle 4$
Since $\angle 3 = \angle 1$,we have $\angle 3 = \angle 4$.
In $\triangle PQC$,since $\angle 3 = \angle 4$,the sides opposite to equal angles are equal,so $PQ = QC$.
Also,tangents drawn from an external point to a circle are equal,so $PQ = BQ$.
Therefore,$BQ = QC$.
Hence,the tangent at $P$ bisects $BC$.
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View Solution32
Easy
$AB$ is a diameter and $AC$ is a chord of a circle with centre $O$ such that $\angle BAC = 30^{\circ}$. The tangent at $C$ intersects the extended $AB$ at a point $D$. Prove that $BC = BD$.
Solution
(N/A) $1$. In $\triangle ABC$,$\angle ACB = 90^{\circ}$ (Angle in a semicircle).
$2$. In $\triangle ABC$,$\angle ABC = 180^{\circ} - (90^{\circ} + 30^{\circ}) = 60^{\circ}$.
$3$. Let $OC$ be the radius. Since $DC$ is a tangent,$OC \perp CD$. Thus,$\angle OCD = 90^{\circ}$.
$4$. $\angle OCA = \angle OAC = 30^{\circ}$ (Since $OA = OC$,$\triangle OAC$ is isosceles).
$5$. $\angle BCD = \angle OCD - \angle OCA = 90^{\circ} - 30^{\circ} = 60^{\circ}$.
$6$. In $\triangle BCD$,$\angle CBD = 180^{\circ} - \angle ABC = 180^{\circ} - 60^{\circ} = 120^{\circ}$ (Linear pair).
$7$. In $\triangle BCD$,$\angle BDC = 180^{\circ} - (120^{\circ} + 60^{\circ}) = 0^{\circ}$? Wait,let us re-evaluate: $\angle BCD = 60^{\circ}$,$\angle CBD = 120^{\circ}$,then $\angle BDC = 180^{\circ} - 180^{\circ} = 0^{\circ}$ is incorrect. Let's use the property: $\angle BCD = \angle BAC = 30^{\circ}$ (Angle between tangent and chord equals angle in alternate segment).
$8$. $\angle BCD = 30^{\circ}$. In $\triangle BCD$,$\angle CBD = 120^{\circ}$,$\angle BCD = 30^{\circ}$,so $\angle BDC = 180^{\circ} - (120^{\circ} + 30^{\circ}) = 30^{\circ}$.
$9$. Since $\angle BCD = \angle BDC = 30^{\circ}$,the triangle $BCD$ is isosceles with $BC = BD$.
$2$. In $\triangle ABC$,$\angle ABC = 180^{\circ} - (90^{\circ} + 30^{\circ}) = 60^{\circ}$.
$3$. Let $OC$ be the radius. Since $DC$ is a tangent,$OC \perp CD$. Thus,$\angle OCD = 90^{\circ}$.
$4$. $\angle OCA = \angle OAC = 30^{\circ}$ (Since $OA = OC$,$\triangle OAC$ is isosceles).
$5$. $\angle BCD = \angle OCD - \angle OCA = 90^{\circ} - 30^{\circ} = 60^{\circ}$.
$6$. In $\triangle BCD$,$\angle CBD = 180^{\circ} - \angle ABC = 180^{\circ} - 60^{\circ} = 120^{\circ}$ (Linear pair).
$7$. In $\triangle BCD$,$\angle BDC = 180^{\circ} - (120^{\circ} + 60^{\circ}) = 0^{\circ}$? Wait,let us re-evaluate: $\angle BCD = 60^{\circ}$,$\angle CBD = 120^{\circ}$,then $\angle BDC = 180^{\circ} - 180^{\circ} = 0^{\circ}$ is incorrect. Let's use the property: $\angle BCD = \angle BAC = 30^{\circ}$ (Angle between tangent and chord equals angle in alternate segment).
$8$. $\angle BCD = 30^{\circ}$. In $\triangle BCD$,$\angle CBD = 120^{\circ}$,$\angle BCD = 30^{\circ}$,so $\angle BDC = 180^{\circ} - (120^{\circ} + 30^{\circ}) = 30^{\circ}$.
$9$. Since $\angle BCD = \angle BDC = 30^{\circ}$,the triangle $BCD$ is isosceles with $BC = BD$.
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View Solution33
Difficult
In the figure,the common tangents $AB$ and $CD$ to two circles with centers $O$ and $O^{\prime}$ intersect at $E$. Prove that the points $O, E, O^{\prime}$ are collinear.
Solution
(N/A) Join $AO, OC$ and $O^{\prime}D, O^{\prime}B$.
In $\triangle E O^{\prime} D$ and $\triangle E O^{\prime} B$:
$O^{\prime} D = O^{\prime} B$ (radii of the same circle)
$O^{\prime} E = O^{\prime} E$ (common side)
$ED = EB$ (tangents drawn from an external point $E$ to the circle are equal in length)
Therefore,$\triangle E O^{\prime} D \cong \triangle E O^{\prime} B$ (by $SSS$ congruence rule).
This implies $\angle O^{\prime} E D = \angle O^{\prime} E B$,so $O^{\prime} E$ is the angle bisector of $\angle DEB$.
Similarly,by considering $\triangle EOA$ and $\triangle EOC$,we can prove $\triangle EOA \cong \triangle EOC$,which implies $\angle OEA = \angle OEC$,so $OE$ is the angle bisector of $\angle AEC$.
Since $AB$ and $CD$ are straight lines intersecting at $E$,$\angle AEC$ and $\angle DEB$ are vertically opposite angles,so $\angle AEC = \angle DEB$.
Since $OE$ bisects $\angle AEC$ and $O^{\prime}E$ bisects $\angle DEB$,and $\angle AEC = \angle DEB$,it follows that $\angle OEC = \angle O^{\prime}EB$.
Since $C, E, B$ lie on a straight line,$\angle OEC + \angle OEO^{\prime} + \angle O^{\prime}EB = 180^{\circ}$.
Substituting $\angle OEC = \angle O^{\prime}EB$,we get $2\angle O^{\prime}EB + \angle OEO^{\prime} = 180^{\circ}$.
However,since $O, E, O^{\prime}$ form a straight line,the angle $\angle OEO^{\prime}$ must be $180^{\circ}$.
Thus,$O, E, O^{\prime}$ are collinear.
In $\triangle E O^{\prime} D$ and $\triangle E O^{\prime} B$:
$O^{\prime} D = O^{\prime} B$ (radii of the same circle)
$O^{\prime} E = O^{\prime} E$ (common side)
$ED = EB$ (tangents drawn from an external point $E$ to the circle are equal in length)
Therefore,$\triangle E O^{\prime} D \cong \triangle E O^{\prime} B$ (by $SSS$ congruence rule).
This implies $\angle O^{\prime} E D = \angle O^{\prime} E B$,so $O^{\prime} E$ is the angle bisector of $\angle DEB$.
Similarly,by considering $\triangle EOA$ and $\triangle EOC$,we can prove $\triangle EOA \cong \triangle EOC$,which implies $\angle OEA = \angle OEC$,so $OE$ is the angle bisector of $\angle AEC$.
Since $AB$ and $CD$ are straight lines intersecting at $E$,$\angle AEC$ and $\angle DEB$ are vertically opposite angles,so $\angle AEC = \angle DEB$.
Since $OE$ bisects $\angle AEC$ and $O^{\prime}E$ bisects $\angle DEB$,and $\angle AEC = \angle DEB$,it follows that $\angle OEC = \angle O^{\prime}EB$.
Since $C, E, B$ lie on a straight line,$\angle OEC + \angle OEO^{\prime} + \angle O^{\prime}EB = 180^{\circ}$.
Substituting $\angle OEC = \angle O^{\prime}EB$,we get $2\angle O^{\prime}EB + \angle OEO^{\prime} = 180^{\circ}$.
However,since $O, E, O^{\prime}$ form a straight line,the angle $\angle OEO^{\prime}$ must be $180^{\circ}$.
Thus,$O, E, O^{\prime}$ are collinear.
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View Solution34
DifficultMCQ
The tangent at a point $C$ of a circle and a diameter $AB$ when extended intersect at $P$. If $\angle PCA = 110^{\circ},$ find $\angle CBA$ [see $Fig.$]. (in $^{\circ}$)
A
$50$
B
$30$
C
$70$
D
$110$
Solution
(C) Let $O$ be the center of the circle. Join $OC$. Since $OC$ is the radius and $PC$ is a tangent at $C,$ $OC \perp PC.$ Therefore,$\angle PCO = 90^{\circ}.$
Given $\angle PCA = 110^{\circ},$ we have $\angle OCA = \angle PCA - \angle PCO = 110^{\circ} - 90^{\circ} = 20^{\circ}.$
In $\triangle OCA,$ $OC = OA$ (radii of the same circle),so $\angle OAC = \angle OCA = 20^{\circ}.$
Since $\angle COA = 180^{\circ} - (20^{\circ} + 20^{\circ}) = 140^{\circ},$ the angle subtended by arc $AC$ at the center is $140^{\circ}.$
The angle subtended by arc $AC$ at the circumference is $\angle ABC = \frac{1}{2} \angle COA = \frac{1}{2} \times 140^{\circ} = 70^{\circ}.$
Thus,$\angle CBA = 70^{\circ}.$
Given $\angle PCA = 110^{\circ},$ we have $\angle OCA = \angle PCA - \angle PCO = 110^{\circ} - 90^{\circ} = 20^{\circ}.$
In $\triangle OCA,$ $OC = OA$ (radii of the same circle),so $\angle OAC = \angle OCA = 20^{\circ}.$
Since $\angle COA = 180^{\circ} - (20^{\circ} + 20^{\circ}) = 140^{\circ},$ the angle subtended by arc $AC$ at the center is $140^{\circ}.$
The angle subtended by arc $AC$ at the circumference is $\angle ABC = \frac{1}{2} \angle COA = \frac{1}{2} \times 140^{\circ} = 70^{\circ}.$
Thus,$\angle CBA = 70^{\circ}.$
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View Solution35
MediumMCQ
$A$ line drawn from the centre $O$ of a circle with radius $7 \, cm$ intersects the tangent at point $P$ on the circle at point $Q$,such that $PQ = 24 \, cm$. Find $OQ$.
A
$12$
B
$25$
C
$28$
D
$32$
Solution
(B) $O$ is the centre of the circle and $P$ is the point of contact of the tangent.
Since the radius is always perpendicular to the tangent at the point of contact,$\angle OPQ = 90^\circ$.
Given: $OP = 7 \, cm$ (radius) and $PQ = 24 \, cm$.
In right-angled $\Delta OPQ$,by Pythagoras theorem:
$OQ^2 = OP^2 + PQ^2$
$OQ^2 = (7)^2 + (24)^2$
$OQ^2 = 49 + 576$
$OQ^2 = 625$
$OQ = \sqrt{625} = 25 \, cm$.
Since the radius is always perpendicular to the tangent at the point of contact,$\angle OPQ = 90^\circ$.
Given: $OP = 7 \, cm$ (radius) and $PQ = 24 \, cm$.
In right-angled $\Delta OPQ$,by Pythagoras theorem:
$OQ^2 = OP^2 + PQ^2$
$OQ^2 = (7)^2 + (24)^2$
$OQ^2 = 49 + 576$
$OQ^2 = 625$
$OQ = \sqrt{625} = 25 \, cm$.
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View Solution36
MediumMCQ
$\odot(P, 3)$ and $\odot(P, 5)$ are two concentric circles. Chord $\overline{AB}$ of $\odot(P, 5)$ touches $\odot(P, 3)$ at $M$. Find $AB$.
A
$6$
B
$18$
C
$8$
D
$12$
Solution
(C) Chord $\overline{AB}$ of $\odot(P, 5)$ touches $\odot(P, 3)$ at $M$.
Since the radius is perpendicular to the tangent at the point of contact,$\overline{PM} \perp \overline{AB}$.
In a circle,a perpendicular drawn from the center to a chord bisects the chord. Therefore,$M$ is the midpoint of $\overline{AB}$.
Given $PM = 3$ (radius of the smaller circle) and $PB = 5$ (radius of the larger circle).
In right-angled $\Delta PMB$,by Pythagoras theorem:
$PB^2 = PM^2 + MB^2$
$5^2 = 3^2 + MB^2$
$25 = 9 + MB^2$
$MB^2 = 25 - 9 = 16$
$MB = \sqrt{16} = 4$.
Since $M$ is the midpoint of $\overline{AB}$,$AB = 2 \times MB = 2 \times 4 = 8$.
Since the radius is perpendicular to the tangent at the point of contact,$\overline{PM} \perp \overline{AB}$.
In a circle,a perpendicular drawn from the center to a chord bisects the chord. Therefore,$M$ is the midpoint of $\overline{AB}$.
Given $PM = 3$ (radius of the smaller circle) and $PB = 5$ (radius of the larger circle).
In right-angled $\Delta PMB$,by Pythagoras theorem:
$PB^2 = PM^2 + MB^2$
$5^2 = 3^2 + MB^2$
$25 = 9 + MB^2$
$MB^2 = 25 - 9 = 16$
$MB = \sqrt{16} = 4$.
Since $M$ is the midpoint of $\overline{AB}$,$AB = 2 \times MB = 2 \times 4 = 8$.
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View Solution37
Medium
Two concentric circles have radii $13$ and $8$. $A$ chord of the larger circle touches the smaller circle. Find the length of the chord.
Solution
(N/A) Let $P$ be the common center of the two concentric circles. Let $\overline{AB}$ be the chord of the larger circle with radius $R = 13$,which touches the smaller circle with radius $r = 8$ at point $M$.
Since $\overline{AB}$ is a tangent to the smaller circle at $M$,the radius $\overline{PM}$ is perpendicular to the chord $\overline{AB}$. Thus,$\angle PMB = 90^{\circ}$.
In the right-angled triangle $\Delta PMB$:
By the Pythagorean theorem,$PB^2 = PM^2 + MB^2$.
Here,$PB = 13$ (radius of the larger circle) and $PM = 8$ (radius of the smaller circle).
$13^2 = 8^2 + MB^2$
$169 = 64 + MB^2$
$MB^2 = 169 - 64 = 105$
$MB = \sqrt{105}$.
Since the perpendicular from the center to a chord bisects the chord,$AB = 2 \times MB$.
$AB = 2\sqrt{105}$.
Since $\overline{AB}$ is a tangent to the smaller circle at $M$,the radius $\overline{PM}$ is perpendicular to the chord $\overline{AB}$. Thus,$\angle PMB = 90^{\circ}$.
In the right-angled triangle $\Delta PMB$:
By the Pythagorean theorem,$PB^2 = PM^2 + MB^2$.
Here,$PB = 13$ (radius of the larger circle) and $PM = 8$ (radius of the smaller circle).
$13^2 = 8^2 + MB^2$
$169 = 64 + MB^2$
$MB^2 = 169 - 64 = 105$
$MB = \sqrt{105}$.
Since the perpendicular from the center to a chord bisects the chord,$AB = 2 \times MB$.
$AB = 2\sqrt{105}$.
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View Solution38
MediumMCQ
Point $A$ lies in the exterior of $\odot(P, 10)$. $A$ line from $A$ touches the circle at $B$. If $PA = 26$,then find the length of $AB$.
A
$24$
B
$42$
C
$32$
D
$23$
Solution
(A) Given that point $A$ is in the exterior of the circle with center $P$ and radius $r = 10$.
Since the line from $A$ touches the circle at $B$,$AB$ is a tangent to the circle.
$A$ radius drawn to the point of tangency is perpendicular to the tangent line. Therefore,$\angle PBA = 90^{\circ}$.
In the right-angled triangle $\triangle PBA$,by the Pythagorean theorem:
$PA^2 = PB^2 + AB^2$.
Given $PA = 26$ and $PB = r = 10$.
$26^2 = 10^2 + AB^2$.
$676 = 100 + AB^2$.
$AB^2 = 676 - 100 = 576$.
$AB = \sqrt{576} = 24$.
Since the line from $A$ touches the circle at $B$,$AB$ is a tangent to the circle.
$A$ radius drawn to the point of tangency is perpendicular to the tangent line. Therefore,$\angle PBA = 90^{\circ}$.
In the right-angled triangle $\triangle PBA$,by the Pythagorean theorem:
$PA^2 = PB^2 + AB^2$.
Given $PA = 26$ and $PB = r = 10$.
$26^2 = 10^2 + AB^2$.
$676 = 100 + AB^2$.
$AB^2 = 676 - 100 = 576$.
$AB = \sqrt{576} = 24$.
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View Solution39
EasyMCQ
Point $A$ lies in the exterior of $\odot(O, 8)$. $A$ line through $A$ touches the circle at $B$. If $AB = 15$,then find $OA$.
A
$23$
B
$17$
C
$32$
D
$15$
Solution
(B) Given that the circle has a center $O$ and radius $r = 8$.
Point $A$ is in the exterior of the circle.
$A$ line through $A$ touches the circle at $B$,which means $AB$ is a tangent to the circle at point $B$.
By the property of tangents,the radius $OB$ is perpendicular to the tangent $AB$ at the point of contact $B$.
Therefore,$\triangle OBA$ is a right-angled triangle with $\angle OBA = 90^{\circ}$.
In $\triangle OBA$,by the Pythagorean theorem:
$OA^2 = OB^2 + AB^2$
Given $OB = r = 8$ and $AB = 15$.
$OA^2 = 8^2 + 15^2$
$OA^2 = 64 + 225$
$OA^2 = 289$
$OA = \sqrt{289} = 17$.
Thus,the length of $OA$ is $17$.
Point $A$ is in the exterior of the circle.
$A$ line through $A$ touches the circle at $B$,which means $AB$ is a tangent to the circle at point $B$.
By the property of tangents,the radius $OB$ is perpendicular to the tangent $AB$ at the point of contact $B$.
Therefore,$\triangle OBA$ is a right-angled triangle with $\angle OBA = 90^{\circ}$.
In $\triangle OBA$,by the Pythagorean theorem:
$OA^2 = OB^2 + AB^2$
Given $OB = r = 8$ and $AB = 15$.
$OA^2 = 8^2 + 15^2$
$OA^2 = 64 + 225$
$OA^2 = 289$
$OA = \sqrt{289} = 17$.
Thus,the length of $OA$ is $17$.
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View Solution40
MediumMCQ
Two concentric circles having radii $41$ and $40$ are given. The chord of the circle with larger radius touches the circle with smaller radius. Find the length of the chord.
A
$9$
B
$18$
C
$81$
D
$82$
Solution
(B) Let the radii of the two concentric circles be $R = 41$ and $r = 40$.
Let $O$ be the common center.
Let $AB$ be the chord of the larger circle that touches the smaller circle at point $P$.
Since the chord touches the smaller circle,$OP$ is perpendicular to $AB$ and $OP = r = 40$.
In the right-angled triangle $\triangle OPA$,by the Pythagorean theorem:
$OA^2 = OP^2 + AP^2$
$41^2 = 40^2 + AP^2$
$1681 = 1600 + AP^2$
$AP^2 = 1681 - 1600 = 81$
$AP = 9$.
Since the perpendicular from the center to a chord bisects the chord,the length of the chord $AB = 2 \times AP = 2 \times 9 = 18$.
Let $O$ be the common center.
Let $AB$ be the chord of the larger circle that touches the smaller circle at point $P$.
Since the chord touches the smaller circle,$OP$ is perpendicular to $AB$ and $OP = r = 40$.
In the right-angled triangle $\triangle OPA$,by the Pythagorean theorem:
$OA^2 = OP^2 + AP^2$
$41^2 = 40^2 + AP^2$
$1681 = 1600 + AP^2$
$AP^2 = 1681 - 1600 = 81$
$AP = 9$.
Since the perpendicular from the center to a chord bisects the chord,the length of the chord $AB = 2 \times AP = 2 \times 9 = 18$.
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View Solution41
MediumMCQ
Point $P$ lies in the exterior of a circle with centre $O$. $OP = 34$ and a tangent through $P$ touches the circle at $Q$. If $PQ = 16$,then find the diameter of the circle.
A
$45$
B
$50$
C
$65$
D
$60$
Solution
(D) In a circle with centre $O$,the radius $OQ$ is perpendicular to the tangent $PQ$ at the point of contact $Q$. Therefore,$\triangle OQP$ is a right-angled triangle with $\angle OQP = 90^{\circ}$.
Using the Pythagorean theorem in $\triangle OQP$:
$OQ^2 + PQ^2 = OP^2$
Given $OP = 34$ and $PQ = 16$:
$OQ^2 + 16^2 = 34^2$
$OQ^2 + 256 = 1156$
$OQ^2 = 1156 - 256 = 900$
$OQ = \sqrt{900} = 30$.
The radius of the circle is $30$. The diameter of the circle is $2 \times \text{radius} = 2 \times 30 = 60$.
Using the Pythagorean theorem in $\triangle OQP$:
$OQ^2 + PQ^2 = OP^2$
Given $OP = 34$ and $PQ = 16$:
$OQ^2 + 16^2 = 34^2$
$OQ^2 + 256 = 1156$
$OQ^2 = 1156 - 256 = 900$
$OQ = \sqrt{900} = 30$.
The radius of the circle is $30$. The diameter of the circle is $2 \times \text{radius} = 2 \times 30 = 60$.
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View Solution42
MediumMCQ
Two concentric circles having radii $17$ and $8$ are given. The chord of the circle with larger radius touches the circle with smaller radius. Find the length of the chord.
A
$30$
B
$25$
C
$35$
D
$40$
Solution
(A) Let the center of the concentric circles be $O$. Let the radius of the larger circle be $R = 17$ and the radius of the smaller circle be $r = 8$.
Let $AB$ be the chord of the larger circle that touches the smaller circle at point $P$.
Since $AB$ is tangent to the smaller circle at $P$,the radius $OP$ is perpendicular to the chord $AB$ $(OP \perp AB)$.
In the right-angled triangle $\triangle OPA$,by the Pythagorean theorem:
$OA^2 = OP^2 + AP^2$
$17^2 = 8^2 + AP^2$
$289 = 64 + AP^2$
$AP^2 = 289 - 64 = 225$
$AP = \sqrt{225} = 15$.
Since the perpendicular from the center to a chord bisects the chord,$AB = 2 \times AP = 2 \times 15 = 30$.
Therefore,the length of the chord is $30$.
Let $AB$ be the chord of the larger circle that touches the smaller circle at point $P$.
Since $AB$ is tangent to the smaller circle at $P$,the radius $OP$ is perpendicular to the chord $AB$ $(OP \perp AB)$.
In the right-angled triangle $\triangle OPA$,by the Pythagorean theorem:
$OA^2 = OP^2 + AP^2$
$17^2 = 8^2 + AP^2$
$289 = 64 + AP^2$
$AP^2 = 289 - 64 = 225$
$AP = \sqrt{225} = 15$.
Since the perpendicular from the center to a chord bisects the chord,$AB = 2 \times AP = 2 \times 15 = 30$.
Therefore,the length of the chord is $30$.
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View Solution43
MediumMCQ
$P$ is a point in the exterior of $\odot(O, r)$ and the tangents from $P$ to the circle touch the circle at $X$ and $Y$. Find $OP$,if $r = 12$ and $XP = 5$.
A
$13$
B
$10$
C
$17$
D
$15$
Solution
(A) Given that $r = OX = 12$ and $XP = 5$.
Since $PX$ is a tangent to the circle at $X$,the radius $OX$ is perpendicular to the tangent $PX$ at the point of contact.
Therefore,$\angle OXP = 90^{\circ}$.
In the right-angled triangle $\Delta OXP$,by the Pythagorean theorem:
$OP^2 = OX^2 + XP^2$
$OP^2 = (12)^2 + (5)^2$
$OP^2 = 144 + 25$
$OP^2 = 169$
$OP = \sqrt{169} = 13$.
Thus,the length of $OP$ is $13$.
Since $PX$ is a tangent to the circle at $X$,the radius $OX$ is perpendicular to the tangent $PX$ at the point of contact.
Therefore,$\angle OXP = 90^{\circ}$.
In the right-angled triangle $\Delta OXP$,by the Pythagorean theorem:
$OP^2 = OX^2 + XP^2$
$OP^2 = (12)^2 + (5)^2$
$OP^2 = 144 + 25$
$OP^2 = 169$
$OP = \sqrt{169} = 13$.
Thus,the length of $OP$ is $13$.
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View Solution44
MediumMCQ
$P$ is a point in the exterior of $\odot(O, r)$ and the tangents from $P$ to the circle touch the circle at $X$ and $Y$. Find $m\angle XOP$,if $m\angle XPO = 65^\circ$. (in $^\circ$)
A
$25$
B
$52$
C
$18$
D
$34$
Solution
(A) $\overline{OX}$ is a radius and $\overleftrightarrow{PX}$ is a tangent at point $X$.
According to the theorem,the tangent at any point of a circle is perpendicular to the radius through the point of contact.
Therefore,$\overline{OX} \perp \overleftrightarrow{PX}$,which implies $m\angle OXP = 90^\circ$.
In the right-angled triangle $\Delta PXO$,the sum of the angles is $180^\circ$.
Therefore,$m\angle OXP + m\angle XPO + m\angle XOP = 180^\circ$.
$90^\circ + 65^\circ + m\angle XOP = 180^\circ$.
$155^\circ + m\angle XOP = 180^\circ$.
$m\angle XOP = 180^\circ - 155^\circ = 25^\circ$.
According to the theorem,the tangent at any point of a circle is perpendicular to the radius through the point of contact.
Therefore,$\overline{OX} \perp \overleftrightarrow{PX}$,which implies $m\angle OXP = 90^\circ$.
In the right-angled triangle $\Delta PXO$,the sum of the angles is $180^\circ$.
Therefore,$m\angle OXP + m\angle XPO + m\angle XOP = 180^\circ$.
$90^\circ + 65^\circ + m\angle XOP = 180^\circ$.
$155^\circ + m\angle XOP = 180^\circ$.
$m\angle XOP = 180^\circ - 155^\circ = 25^\circ$.
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View Solution45
DifficultMCQ
Radii of two concentric circles are $13$ and $8$. $\overline{AB}$ is a diameter of the circle with the larger radius. $\overline{BD}$ touches the circle with the smaller radius at $D$. Find $AD$.
A
$29$
B
$23$
C
$17$
D
$19$
Solution
(D) Let the center of the concentric circles be $P$. The radii are $R = 13$ and $r = 8$.
$\overline{AB}$ is a diameter of the larger circle,so $AB = 2 \times 13 = 26$ and $PB = 13$.
$\overline{BD}$ is tangent to the smaller circle at $D$,so $\overline{PD} \perp \overline{BD}$ and $PD = 8$.
In right-angled $\Delta PDB$,by the Pythagorean theorem:
$BD^2 = PB^2 - PD^2 = 13^2 - 8^2 = 169 - 64 = 105$.
Let $\overline{BD}$ intersect the larger circle at $C$. Since $\overline{PD} \perp \overline{BC}$,$D$ is the midpoint of chord $\overline{BC}$,so $CD = BD$. Thus,$CD^2 = 105$.
In the larger circle,$\angle ACB$ is an angle in a semicircle,so $\angle ACB = 90^\circ$.
In $\Delta ABC$ and $\Delta PBD$,$\angle ACB = \angle PDB = 90^\circ$ and $\angle B$ is common. Thus,$\Delta ABC \sim \Delta PBD$ by $AA$ similarity.
Therefore,$\frac{AB}{PB} = \frac{AC}{PD} \implies \frac{26}{13} = \frac{AC}{8} \implies 2 = \frac{AC}{8} \implies AC = 16$.
In right-angled $\Delta ACD$,by the Pythagorean theorem:
$AD^2 = AC^2 + CD^2 = 16^2 + 105 = 256 + 105 = 361$.
$AD = \sqrt{361} = 19$.
$\overline{AB}$ is a diameter of the larger circle,so $AB = 2 \times 13 = 26$ and $PB = 13$.
$\overline{BD}$ is tangent to the smaller circle at $D$,so $\overline{PD} \perp \overline{BD}$ and $PD = 8$.
In right-angled $\Delta PDB$,by the Pythagorean theorem:
$BD^2 = PB^2 - PD^2 = 13^2 - 8^2 = 169 - 64 = 105$.
Let $\overline{BD}$ intersect the larger circle at $C$. Since $\overline{PD} \perp \overline{BC}$,$D$ is the midpoint of chord $\overline{BC}$,so $CD = BD$. Thus,$CD^2 = 105$.
In the larger circle,$\angle ACB$ is an angle in a semicircle,so $\angle ACB = 90^\circ$.
In $\Delta ABC$ and $\Delta PBD$,$\angle ACB = \angle PDB = 90^\circ$ and $\angle B$ is common. Thus,$\Delta ABC \sim \Delta PBD$ by $AA$ similarity.
Therefore,$\frac{AB}{PB} = \frac{AC}{PD} \implies \frac{26}{13} = \frac{AC}{8} \implies 2 = \frac{AC}{8} \implies AC = 16$.
In right-angled $\Delta ACD$,by the Pythagorean theorem:
$AD^2 = AC^2 + CD^2 = 16^2 + 105 = 256 + 105 = 361$.
$AD = \sqrt{361} = 19$.
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View Solution46
Difficult
$\overline{AB}$ is a chord of a circle with centre $O$. Line $l$ touches the circle at $B$. The foot of the perpendicular from $A$ to $l$ is $D$. Prove that $\angle BAO \cong \angle BAD$.
Solution
(N/A) Given: $\overline{AB}$ is a chord of a circle with centre $O$. Line $l$ is a tangent to the circle at $B$. $D$ is the foot of the perpendicular from $A$ to line $l$ (i.e.,$\overline{AD} \perp l$).
To prove: $\angle BAO \cong \angle BAD$.
Proof:
$1$. In $\Delta OAB$,$\overline{OA} \cong \overline{OB}$ (radii of the same circle).
$2$. Therefore,$\angle ABO \cong \angle BAO$ (angles opposite to equal sides are equal) ... $(1)$.
$3$. Line $l$ is a tangent to the circle at $B$,and $\overline{OB}$ is the radius through the point of contact $B$. Therefore,$\overline{OB} \perp l$.
$4$. We are given that $\overline{AD} \perp l$.
$5$. Since both $\overline{OB}$ and $\overline{AD}$ are perpendicular to the same line $l$,they must be parallel to each other $(\overline{OB} \parallel \overline{AD})$.
$6$. Considering $\overline{OB} \parallel \overline{AD}$ with $\overleftrightarrow{AB}$ as a transversal,the alternate interior angles are equal.
$7$. Therefore,$\angle ABO \cong \angle BAD$ (alternate interior angles) ... $(2)$.
$8$. From equations $(1)$ and $(2)$,since both $\angle BAO$ and $\angle BAD$ are equal to $\angle ABO$,we conclude that $\angle BAO \cong \angle BAD$.
To prove: $\angle BAO \cong \angle BAD$.
Proof:
$1$. In $\Delta OAB$,$\overline{OA} \cong \overline{OB}$ (radii of the same circle).
$2$. Therefore,$\angle ABO \cong \angle BAO$ (angles opposite to equal sides are equal) ... $(1)$.
$3$. Line $l$ is a tangent to the circle at $B$,and $\overline{OB}$ is the radius through the point of contact $B$. Therefore,$\overline{OB} \perp l$.
$4$. We are given that $\overline{AD} \perp l$.
$5$. Since both $\overline{OB}$ and $\overline{AD}$ are perpendicular to the same line $l$,they must be parallel to each other $(\overline{OB} \parallel \overline{AD})$.
$6$. Considering $\overline{OB} \parallel \overline{AD}$ with $\overleftrightarrow{AB}$ as a transversal,the alternate interior angles are equal.
$7$. Therefore,$\angle ABO \cong \angle BAD$ (alternate interior angles) ... $(2)$.
$8$. From equations $(1)$ and $(2)$,since both $\angle BAO$ and $\angle BAD$ are equal to $\angle ABO$,we conclude that $\angle BAO \cong \angle BAD$.
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View Solution47
Difficult
$A$ circle with centre $O$ touches sides $\overline{AB}$,$\overline{BC}$,$\overline{CD}$ and $\overline{DA}$ of quadrilateral $ABCD$ at points $P, Q, R$ and $S$ respectively. Prove that $m \angle AOB + m \angle COD = 180^{\circ}$ and $m \angle AOD + m \angle BOC = 180^{\circ}$.
Solution
(N/A) circle with centre $O$ touches the sides $\overline{AB}$,$\overline{BC}$,$\overline{CD}$ and $\overline{DA}$ of quadrilateral $ABCD$ at $P, Q, R$ and $S$ respectively.
To prove: $m \angle AOB + m \angle COD = 180^{\circ}$ and $m \angle AOD + m \angle BOC = 180^{\circ}$.
Proof: Draw $\overline{OP}, \overline{OQ}, \overline{OR}$ and $\overline{OS}$.
Since tangents from an external point to a circle are equal,$\overline{AS} \cong \overline{AP}$.
In $\Delta ASO$ and $\Delta APO$:
$\overline{AS} \cong \overline{AP}$ (Tangents from point $A$)
$\overline{OS} \cong \overline{OP}$ (Radii of the same circle)
$\overline{AO} \cong \overline{AO}$ (Common side)
Therefore,$\Delta ASO \cong \Delta APO$ ($SSS$ congruence rule).
This implies $m \angle OAS = m \angle OAP$,so $m \angle OAB = \frac{1}{2} m \angle DAB$.
Similarly,we can show:
$m \angle OBA = \frac{1}{2} m \angle ABC$
$m \angle OCD = \frac{1}{2} m \angle BCD$
$m \angle ODC = \frac{1}{2} m \angle CDA$
In $\Delta AOB$,$m \angle AOB = 180^{\circ} - (m \angle OAB + m \angle OBA) = 180^{\circ} - \frac{1}{2}(m \angle DAB + m \angle ABC)$.
In $\Delta COD$,$m \angle COD = 180^{\circ} - (m \angle OCD + m \angle ODC) = 180^{\circ} - \frac{1}{2}(m \angle BCD + m \angle CDA)$.
Adding these:
$m \angle AOB + m \angle COD = 360^{\circ} - \frac{1}{2}(m \angle DAB + m \angle ABC + m \angle BCD + m \angle CDA)$.
Since the sum of angles in a quadrilateral is $360^{\circ}$:
$m \angle AOB + m \angle COD = 360^{\circ} - \frac{1}{2}(360^{\circ}) = 360^{\circ} - 180^{\circ} = 180^{\circ}$.
Similarly,it can be proved that $m \angle AOD + m \angle BOC = 180^{\circ}$.
To prove: $m \angle AOB + m \angle COD = 180^{\circ}$ and $m \angle AOD + m \angle BOC = 180^{\circ}$.
Proof: Draw $\overline{OP}, \overline{OQ}, \overline{OR}$ and $\overline{OS}$.
Since tangents from an external point to a circle are equal,$\overline{AS} \cong \overline{AP}$.
In $\Delta ASO$ and $\Delta APO$:
$\overline{AS} \cong \overline{AP}$ (Tangents from point $A$)
$\overline{OS} \cong \overline{OP}$ (Radii of the same circle)
$\overline{AO} \cong \overline{AO}$ (Common side)
Therefore,$\Delta ASO \cong \Delta APO$ ($SSS$ congruence rule).
This implies $m \angle OAS = m \angle OAP$,so $m \angle OAB = \frac{1}{2} m \angle DAB$.
Similarly,we can show:
$m \angle OBA = \frac{1}{2} m \angle ABC$
$m \angle OCD = \frac{1}{2} m \angle BCD$
$m \angle ODC = \frac{1}{2} m \angle CDA$
In $\Delta AOB$,$m \angle AOB = 180^{\circ} - (m \angle OAB + m \angle OBA) = 180^{\circ} - \frac{1}{2}(m \angle DAB + m \angle ABC)$.
In $\Delta COD$,$m \angle COD = 180^{\circ} - (m \angle OCD + m \angle ODC) = 180^{\circ} - \frac{1}{2}(m \angle BCD + m \angle CDA)$.
Adding these:
$m \angle AOB + m \angle COD = 360^{\circ} - \frac{1}{2}(m \angle DAB + m \angle ABC + m \angle BCD + m \angle CDA)$.
Since the sum of angles in a quadrilateral is $360^{\circ}$:
$m \angle AOB + m \angle COD = 360^{\circ} - \frac{1}{2}(360^{\circ}) = 360^{\circ} - 180^{\circ} = 180^{\circ}$.
Similarly,it can be proved that $m \angle AOD + m \angle BOC = 180^{\circ}$.
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View Solution48
MediumMCQ
$A$ line passing through the centre of the circle $O$ intersects the tangent to the circle at $Q$. $P$ is the point of contact. If the radius of the circle is $9$ and $PQ = 40$,then find $OQ$.
A
$41$
B
$48$
C
$55$
D
$31$
Solution
(A) The radius of the circle is $OP = 9$.
Since $PQ$ is a tangent to the circle at point $P$,the radius $OP$ is perpendicular to the tangent $PQ$ at the point of contact.
Therefore,$\angle OPQ = 90^{\circ}$.
In the right-angled triangle $\triangle OPQ$,by the Pythagorean theorem,we have:
$OQ^2 = OP^2 + PQ^2$.
Substituting the given values:
$OQ^2 = 9^2 + 40^2$.
$OQ^2 = 81 + 1600$.
$OQ^2 = 1681$.
Taking the square root of both sides:
$OQ = \sqrt{1681} = 41$.
Thus,the length of $OQ$ is $41$.
Since $PQ$ is a tangent to the circle at point $P$,the radius $OP$ is perpendicular to the tangent $PQ$ at the point of contact.
Therefore,$\angle OPQ = 90^{\circ}$.
In the right-angled triangle $\triangle OPQ$,by the Pythagorean theorem,we have:
$OQ^2 = OP^2 + PQ^2$.
Substituting the given values:
$OQ^2 = 9^2 + 40^2$.
$OQ^2 = 81 + 1600$.
$OQ^2 = 1681$.
Taking the square root of both sides:
$OQ = \sqrt{1681} = 41$.
Thus,the length of $OQ$ is $41$.
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View Solution49
MediumMCQ
In $\Delta ABC$,$m \angle B = 90^{\circ}$. $A$ circle is inscribed in $\Delta ABC$ touching all its sides. If $AB = 16$ and $BC = 30$,find the radius of the circle.
A
$8$
B
$6$
C
$12$
D
$18$
Solution
(B) In a right-angled triangle,the radius $r$ of the incircle is given by the formula $r = \frac{AB + BC - AC}{2}$.
First,calculate the hypotenuse $AC$ using the Pythagorean theorem: $AC = \sqrt{AB^2 + BC^2} = \sqrt{16^2 + 30^2} = \sqrt{256 + 900} = \sqrt{1156} = 34$.
Now,substitute the values into the formula: $r = \frac{16 + 30 - 34}{2}$.
$r = \frac{46 - 34}{2} = \frac{12}{2} = 6$.
Thus,the radius of the circle is $6$.
First,calculate the hypotenuse $AC$ using the Pythagorean theorem: $AC = \sqrt{AB^2 + BC^2} = \sqrt{16^2 + 30^2} = \sqrt{256 + 900} = \sqrt{1156} = 34$.
Now,substitute the values into the formula: $r = \frac{16 + 30 - 34}{2}$.
$r = \frac{46 - 34}{2} = \frac{12}{2} = 6$.
Thus,the radius of the circle is $6$.
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View Solution50
DifficultMCQ
Point $P$ lies in the exterior of $\odot(O, r)$. $\overleftrightarrow{PQ}$ touches the circle at $Q$. If $PO = 26$ and $PQ = 10$,then find the diameter of the circle.
A
$67$
B
$54$
C
$48$
D
$34$
Solution
(C) Given that $P$ is a point outside the circle with center $O$ and radius $r$. $\overleftrightarrow{PQ}$ is a tangent to the circle at point $Q$.
By the property of tangents,the radius $OQ$ is perpendicular to the tangent $PQ$ at the point of contact $Q$. Thus,$\triangle OQP$ is a right-angled triangle with $\angle OQP = 90^{\circ}$.
In $\triangle OQP$,by the Pythagorean theorem:
$OP^2 = OQ^2 + PQ^2$
Given $OP = 26$ and $PQ = 10$,we have:
$26^2 = r^2 + 10^2$
$676 = r^2 + 100$
$r^2 = 676 - 100 = 576$
$r = \sqrt{576} = 24$.
The diameter of the circle is $d = 2r = 2 \times 24 = 48$.
By the property of tangents,the radius $OQ$ is perpendicular to the tangent $PQ$ at the point of contact $Q$. Thus,$\triangle OQP$ is a right-angled triangle with $\angle OQP = 90^{\circ}$.
In $\triangle OQP$,by the Pythagorean theorem:
$OP^2 = OQ^2 + PQ^2$
Given $OP = 26$ and $PQ = 10$,we have:
$26^2 = r^2 + 10^2$
$676 = r^2 + 100$
$r^2 = 676 - 100 = 576$
$r = \sqrt{576} = 24$.
The diameter of the circle is $d = 2r = 2 \times 24 = 48$.
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View SolutionCircles — Tangent to a Circle · Frequently Asked Questions
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