Perpendicular from the Centre to a Chord Questions in English

(N/A) Steps of construction:
$I.$ Take any three points on the given circle. Let these be $A, B$ and $C$.
$II.$ Join $AB$ and $BC$.
$III.$ Draw the perpendicular bisector $PQ$ of $AB$.
$IV.$ Draw the perpendicular bisector $RS$ of $BC$ such that it intersects $PQ$ at $O$.
Thus,$O$ is the required centre of the given circle.

(N/A) Let there be two circles with centres $O$ and $O^{\prime}$ that intersect at points $A$ and $B$.
$AB$ is the common chord of the two circles,and $OO^{\prime}$ is the line segment joining the centres. Let $OO^{\prime}$ and $AB$ intersect at point $M$.
To prove that $OO^{\prime}$ is the perpendicular bisector of $AB$,we join $OA, OB, O^{\prime}A$,and $O^{\prime}B$.
In $\Delta OAO^{\prime}$ and $\Delta OBO^{\prime}$:
$OA = OB$ (radii of the same circle)
$O^{\prime}A = O^{\prime}B$ (radii of the same circle)
$OO^{\prime} = OO^{\prime}$ (common side)
By $SSS$ congruence criterion,$\Delta OAO^{\prime} \cong \Delta OBO^{\prime}$.
Therefore,$\angle 1 = \angle 2$ (by $CPCT$).
Now,in $\Delta AOM$ and $\Delta BOM$:
$OA = OB$ (radii of the same circle)
$OM = OM$ (common side)
$\angle 1 = \angle 2$ (proved above)
By $SAS$ congruence criterion,$\Delta AOM \cong \Delta BOM$.
Therefore,$AM = BM$ (by $CPCT$) and $\angle 3 = \angle 4$ (by $CPCT$).
Since $AB$ is a line segment,$\angle 3 + \angle 4 = 180^{\circ}$ (linear pair).
Thus,$2 \angle 3 = 180^{\circ} \Rightarrow \angle 3 = 90^{\circ}$.
Since $\angle 3 = \angle 4 = 90^{\circ}$ and $AM = BM$,$OO^{\prime}$ is the perpendicular bisector of $AB$.

(N/A) We have two circles with the common centre $O$.
$A$ line $\ell$ intersects the outer circle at $A$ and $D$ and the inner circle at $B$ and $C$. To prove that $AB = CD$,let us draw $OM \perp \ell$.
For the outer circle,
$\because OM \perp \ell$,and the perpendicular from the centre to a chord bisects the chord,
$\therefore AM = MD$ --- $(1)$
For the inner circle,
$\because OM \perp \ell$,
$\therefore BM = MC$ --- $(2)$
Subtracting $(2)$ from $(1)$,we have:
$AM - BM = MD - MC$
$AB = CD$
Hence,it is proved.

(C) Let $O$ be the centre of the circle. Draw $OP \perp AB$.
Since the perpendicular from the centre to a chord bisects the chord,we have:
$AP = \frac{1}{2} \times AB = \frac{1}{2} \times 30 = 15 \, cm$.
The radius of the circle is $OA = \frac{AD}{2} = \frac{34}{2} = 17 \, cm$.
In the right-angled triangle $\Delta OPA$,by the Pythagorean theorem:
$OP^2 + AP^2 = OA^2$
$OP^2 + 15^2 = 17^2$
$OP^2 + 225 = 289$
$OP^2 = 289 - 225 = 64$
$OP = \sqrt{64} = 8 \, cm$.
Thus,the distance of the chord $AB$ from the centre is $8 \, cm$.

(D) Since the perpendicular from the centre of a circle to a chord bisects the chord,we have:
$AC = \frac{1}{2} \times AB = \frac{1}{2} \times 8 \text{ cm} = 4 \text{ cm}$.
In the right-angled triangle $\triangle OCA$,by the Pythagoras theorem:
$OC^2 + AC^2 = OA^2$
$OC^2 + 4^2 = 5^2$
$OC^2 + 16 = 25$
$OC^2 = 25 - 16 = 9$
$OC = \sqrt{9} = 3 \text{ cm}$.
Since $OD$ is the radius of the circle,$OD = OA = 5 \text{ cm}$.
Therefore,$CD = OD - OC = 5 \text{ cm} - 3 \text{ cm} = 2 \text{ cm}$.

(N/A) Given: $PQ$ is the perpendicular bisector of chord $AB$ of a circle,intersecting the circle at $P$ and $Q$. Let $M$ be the point of intersection of $PQ$ and $AB$.
To prove: Arc $PXA \cong$ Arc $PYB$.
Proof:
In $\Delta APM$ and $\Delta BPM$:
$1$. $AM = BM$ (Since $PQ$ is the perpendicular bisector of $AB$)
$2$. $\angle AMP = \angle BMP = 90^{\circ}$ (Given)
$3$. $PM = PM$ (Common side)
Therefore,$\Delta APM \cong \Delta BPM$ by the $SAS$ congruence rule.
This implies $AP = BP$ by $CPCT$.
Since $AP$ and $BP$ are chords of the circle and $AP = BP$,their corresponding arcs are congruent.
Thus,arc $PXA \cong$ arc $PYB$.

(N/A) Given: $AB$ and $CD$ are two chords of a circle with centre $O$. The mid-points of $AB$ and $CD$ are $L$ and $M$ respectively.
To prove: $AB \parallel CD$
Proof: Since $L$ is the mid-point of chord $AB$,therefore $OL \perp AB$,or $\angle ALO = 90^{\circ}$.
[Because the line joining the centre of a circle to the mid-point of a chord is perpendicular to the chord]
Similarly,$\angle CMO = 90^{\circ}$.
Therefore,$\angle ALO = \angle CMO = 90^{\circ}$.
Since these are corresponding angles formed by the transversal $LM$ intersecting lines $AB$ and $CD$,and they are equal,the lines must be parallel.
So,$AB \parallel CD$.
Hence,proved.

(N/A) Given: Two circles with centres $O$ and $O'$ intersect at $A$ and $B$. $A$ line $PQ$ is drawn parallel to $OO'$ passing through $A$,intersecting the circles at $P$ and $Q$.
To prove: $PQ = 2 OO'$.
Construction: Draw $OC \perp PA$ and $O'D \perp AQ$.
Proof:
$1$. Since the perpendicular from the centre of a circle to a chord bisects the chord,we have:
$PA = 2 CA$ (as $OC \perp PA$) $...(1)$
$AQ = 2 AD$ (as $O'D \perp AQ$) $...(2)$
$2$. Adding equations $(1)$ and $(2)$:
$PA + AQ = 2 CA + 2 AD$
$PQ = 2(CA + AD)$
$3$. Since $PQ \parallel OO'$,$OC \perp PQ$,and $O'D \perp PQ$,the quadrilateral $CDO'O$ is a rectangle.
Therefore,$CD = OO'$.
$4$. Substituting $CD = OO'$ in the equation $PQ = 2(CA + AD)$:
$PQ = 2 CD = 2 OO'$.
Hence,$PQ = 2 OO'$ is proved.

(N/A) Let $P$ be the given point inside a circle with centre $O$. Draw the chord $AB$ which is perpendicular to the diameter $XY$ passing through $P$. Let $CD$ be any other chord passing through $P$. Draw $ON$ perpendicular to $CD$ from $O$.
In $\Delta ONP$,$\angle ONP = 90^{\circ}$. Since $OP$ is the hypotenuse,$OP > ON$.
We know that the chord nearer to the centre is larger than the chord which is farther from the centre.
Since $ON < OP$ (where $OP$ is the distance of chord $AB$ from the centre),the chord $CD$ is farther from the centre than the chord $AB$.
Therefore,$CD > AB$.
In other words,$AB$ is the smallest of all chords passing through $P$.

(N/A) We have a circle with centre $O$. $BC$ is a diameter and $AB$ is a chord such that $OD \perp AB$. Join $AC$.
The perpendicular from the centre of a circle to a chord bisects the chord. Therefore,$D$ is the midpoint of $AB$.
Since $O$ is the centre of the circle,$O$ is the midpoint of the diameter $BC$.
In $\triangle ABC$,$O$ is the midpoint of $BC$ and $D$ is the midpoint of $AB$. Therefore,$OD$ is the line segment joining the midpoints of two sides of $\triangle ABC$.
By the Midpoint Theorem,the line segment joining the midpoints of two sides of a triangle is parallel to the third side and is half of it.
Therefore,$OD \parallel CA$ and $OD = \frac{1}{2} CA$.
Hence,$CA = 2OD$.

(N/A) $1$. Let the circle have center $P$. Since $l$ is the perpendicular bisector of chord $AB$,it must pass through the center $P$ of the circle.
$2$. We are given that $AB \parallel CD$.
$3$. Since $l$ is perpendicular to $AB$ $(l \perp AB)$ and $AB \parallel CD$,it follows that $l$ must also be perpendicular to $CD$ $(l \perp CD)$.
$4$. $A$ line passing through the center of a circle and perpendicular to a chord bisects the chord.
$5$. Since $l$ passes through the center $P$ and is perpendicular to $CD$,$l$ must bisect chord $CD$.

(N/A) $1$. Let the circle have centre $P$. Let the line $l$ pass through $P$ and intersect $AB$ at point $M$ and $XY$ at point $N$.
$2$. Since the line $l$ bisects chord $AB$ at $M$,by the theorem 'the line drawn through the centre of a circle to bisect a chord is perpendicular to the chord',we have $PM \perp AB$. Thus,$\angle PMA = 90^{\circ}$.
$3$. Similarly,since the line $l$ bisects chord $XY$ at $N$,we have $PN \perp XY$. Thus,$\angle PNY = 90^{\circ}$.
$4$. Since $M, P,$ and $N$ lie on the same line $l$,the angles $\angle PMA$ and $\angle PNY$ are corresponding angles formed by the transversal $l$ intersecting lines $AB$ and $XY$.
$5$. Since $\angle PMA = 90^{\circ}$ and $\angle PNY = 90^{\circ}$,the corresponding angles are equal $(\angle PMA = \angle PNY = 90^{\circ})$.
$6$. Therefore,by the converse of the corresponding angles axiom,$AB \parallel XY$.

(C) From $P$,draw $PM \perp AB$. Join $PA$.
The perpendicular from the centre of a circle to a chord bisects the chord.
$\therefore AM = MB = \frac{1}{2} AB = \frac{1}{2} \times 32 = 16\, cm$.
$PA$ is the radius of the circle,so $PA = 20\, cm$.
In $\Delta PMA$,$\angle M = 90^{\circ}$.
By Pythagoras theorem,$PA^2 = PM^2 + AM^2$.
$\therefore PM^2 = PA^2 - AM^2 = (20)^2 - (16)^2 = 400 - 256 = 144$.
$\therefore PM = \sqrt{144} = 12\, cm$.
Thus,the distance of the chord $AB$ from the centre $P$ is $12\, cm$.

(B) Let $P$ be the centre of the circle and $AB$ be the chord of length $30\,cm$.
Draw a perpendicular $PM$ from the centre $P$ to the chord $AB$.
$A$ perpendicular from the centre of a circle to a chord bisects the chord.
Therefore,$AM = MB = \frac{AB}{2} = \frac{30}{2} = 15\,cm$.
In the right-angled triangle $\triangle PMA$,by Pythagoras theorem:
$PA^2 = PM^2 + AM^2$.
Given radius $PA = 17\,cm$ and $AM = 15\,cm$.
$17^2 = PM^2 + 15^2$.
$289 = PM^2 + 225$.
$PM^2 = 289 - 225 = 64$.
$PM = \sqrt{64} = 8\,cm$.
Thus,the distance of the chord $AB$ from the centre $P$ is $8\,cm$.

(A) Let $P$ be the centre of the circle and $AB$ be the chord of length $40 \, cm$.
Let $M$ be the midpoint of $AB$. The perpendicular from the centre to a chord bisects the chord.
Therefore,$AM = MB = \frac{40}{2} = 20 \, cm$.
Given the distance from the centre to the chord is $PM = 21 \, cm$.
In the right-angled triangle $\triangle PMA$,by the Pythagorean theorem:
$PA^2 = PM^2 + AM^2$
$PA^2 = 21^2 + 20^2$
$PA^2 = 441 + 400 = 841$
$PA = \sqrt{841} = 29 \, cm$.
Here,$PA$ is the radius $(r)$ of the circle.
Diameter $(d)$ = $2 \times r = 2 \times 29 = 58 \, cm$.

(A) Let $P$ be the centre of the circle with radius $r = 25\,cm$.
Let $M$ and $N$ be the midpoints of chords $AB$ and $CD$ respectively.
Since the perpendicular from the centre to a chord bisects the chord,$AM = MB = AB/2 = 40/2 = 20\,cm$ and $CN = ND = CD/2 = 30/2 = 15\,cm$.
In right-angled $\triangle PMA$,by Pythagoras theorem: $PM^2 + AM^2 = PA^2 \implies PM^2 + 20^2 = 25^2 \implies PM^2 = 625 - 400 = 225 \implies PM = 15\,cm$.
In right-angled $\triangle PNC$,by Pythagoras theorem: $PN^2 + CN^2 = PC^2 \implies PN^2 + 15^2 = 25^2 \implies PN^2 = 625 - 225 = 400 \implies PN = 20\,cm$.
Since the centre $P$ is not between the chords,the distance between the chords is $|PN - PM| = |20 - 15| = 5\,cm$.

(C) Let $P$ be the centre of the circle and $r = 26\,cm$ be the radius.
Draw perpendiculars $PM$ and $PN$ from $P$ to chords $AB$ and $CD$ respectively.
Since the perpendicular from the centre to a chord bisects the chord,we have:
$AM = MB = \frac{AB}{2} = \frac{20}{2} = 10\,cm$
$CN = ND = \frac{CD}{2} = \frac{48}{2} = 24\,cm$
In right-angled $\triangle PMA$,by Pythagoras theorem:
$PM^2 + AM^2 = PA^2$
$PM^2 + 10^2 = 26^2$
$PM^2 = 676 - 100 = 576$
$PM = \sqrt{576} = 24\,cm$
In right-angled $\triangle PNC$,by Pythagoras theorem:
$PN^2 + CN^2 = PC^2$
$PN^2 + 24^2 = 26^2$
$PN^2 = 676 - 576 = 100$
$PN = \sqrt{100} = 10\,cm$
Since the centre $P$ lies between the chords,the distance between $AB$ and $CD$ is $PM + PN = 24 + 10 = 34\,cm$.

(A) Let $AB$ and $CD$ be two chords of a circle with centre $O$. Let $M$ and $N$ be the midpoints of chords $AB$ and $CD$ respectively.
Given that the line $MN$ passes through the centre $O$.
Since $M$ is the midpoint of $AB$,the line segment joining the centre to the midpoint of a chord is perpendicular to the chord. Therefore,$OM \perp AB$,which implies $\angle OMA = 90^{\circ}$.
Similarly,since $N$ is the midpoint of $CD$,$ON \perp CD$,which implies $\angle ONC = 90^{\circ}$.
Since the points $M, O, N$ are collinear (as the line $MN$ passes through the centre $O$),the line $MN$ acts as a transversal to the chords $AB$ and $CD$.
The angles $\angle OMA$ and $\angle ONC$ are corresponding angles (or interior angles on the same side depending on the orientation). Specifically,$\angle OMA + \angle ONC = 90^{\circ} + 90^{\circ} = 180^{\circ}$ if they are interior angles on the same side.
Since the sum of the interior angles on the same side of the transversal $MN$ is $180^{\circ}$,the lines $AB$ and $CD$ must be parallel.

(B) Given that the diameter of the circle is $20 \, cm$,the radius $r = \frac{20}{2} = 10 \, cm$.
Let $M$ be the midpoint of the chord $AB$ such that $PM \perp AB$. The distance from the centre $P$ to the chord $AB$ is $PM = 6 \, cm$.
In the right-angled triangle $\triangle PMA$,by the Pythagorean theorem:
$PA^2 = PM^2 + AM^2$
$10^2 = 6^2 + AM^2$
$100 = 36 + AM^2$
$AM^2 = 100 - 36 = 64$
$AM = \sqrt{64} = 8 \, cm$.
Since the perpendicular from the centre to a chord bisects the chord,$AB = 2 \times AM = 2 \times 8 = 16 \, cm$.

(A) Let $P$ be the centre of the circle and $AB$ be the chord. Let $M$ be the midpoint of the chord $AB$ such that $PM \perp AB$.
Given that the distance of the chord from the centre is $PM = 20 \, cm$.
The length of the chord $AB = 96 \, cm$.
Since the perpendicular from the centre to a chord bisects the chord,we have $AM = MB = \frac{AB}{2} = \frac{96}{2} = 48 \, cm$.
In the right-angled triangle $\triangle PMA$,by the Pythagorean theorem,$PA^2 = PM^2 + AM^2$.
Here,$PA$ is the radius $(r)$ of the circle.
$r^2 = 20^2 + 48^2$.
$r^2 = 400 + 2304 = 2704$.
$r = \sqrt{2704} = 52 \, cm$.
Thus,the radius of the circle is $52 \, cm$.

(B) Given: Diameter of the circle $d = 70 \, cm$.
Radius $r = \frac{d}{2} = \frac{70}{2} = 35 \, cm$.
Let $M$ be the midpoint of chord $AB$ such that $PM \perp AB$.
The distance from the centre to the chord is $PM = 21 \, cm$.
In right-angled triangle $\triangle PMA$,by Pythagoras theorem:
$PA^2 = PM^2 + AM^2$
$35^2 = 21^2 + AM^2$
$1225 = 441 + AM^2$
$AM^2 = 1225 - 441 = 784$
$AM = \sqrt{784} = 28 \, cm$.
Since the perpendicular from the centre to a chord bisects the chord,$AB = 2 \times AM = 2 \times 28 = 56 \, cm$.

(B) line segment joining any two points on a circle is known as a chord of the circle. If this chord passes through the center of the circle,it is called the diameter,which is the longest chord of the circle.

(D) The theorem states that a perpendicular drawn from the centre of a circle to a chord bisects the chord. Therefore,the correct option is $D$.

(A) Let the circle have centre $O$ and radius $r = 17 \, cm$. Let $AB$ be the chord of length $30 \, cm$.
Draw a perpendicular $OM$ from the centre $O$ to the chord $AB$.
The perpendicular from the centre of a circle to a chord bisects the chord.
Therefore,$AM = MB = \frac{AB}{2} = \frac{30}{2} = 15 \, cm$.
In the right-angled triangle $\triangle OMA$,by the Pythagorean theorem:
$OA^2 = OM^2 + AM^2$.
Substituting the values: $17^2 = OM^2 + 15^2$.
$289 = OM^2 + 225$.
$OM^2 = 289 - 225 = 64$.
$OM = \sqrt{64} = 8 \, cm$.
Thus,the distance of the chord from the centre is $8 \, cm$.

(B) Let the circle have centre $O$ and radius $r = 13 \, cm$. Let $AB$ be the chord at a distance $OM = 12 \, cm$ from the centre,where $OM \perp AB$.
In the right-angled triangle $\triangle OMA$,by the Pythagorean theorem:
$OA^2 = OM^2 + AM^2$
$13^2 = 12^2 + AM^2$
$169 = 144 + AM^2$
$AM^2 = 169 - 144 = 25$
$AM = \sqrt{25} = 5 \, cm$.
Since the perpendicular from the centre to a chord bisects the chord,the length of the chord $AB = 2 \times AM = 2 \times 5 = 10 \, cm$.