Mix Examples - Circles Questions in English

(C) According to the alternate segment theorem,the angle between a chord and the tangent through one of the endpoints of the chord is equal to the angle in the alternate segment.
Therefore,$\angle PRQ = \angle QPT$.
Given that $\angle QPT = 60^{\circ}$.
Thus,$\angle PRQ = 60^{\circ}$.
Wait,let's re-evaluate the geometry. The angle $\angle PRQ$ is in the alternate segment. If $R$ is on the major arc,then $\angle PRQ = \angle QPT = 60^{\circ}$. However,looking at the figure,$R$ is on the major arc. If the question implies the angle subtended by the chord in the alternate segment,it is $60^{\circ}$. If $R$ is on the minor arc,the angle would be $180^{\circ} - 60^{\circ} = 120^{\circ}$. Given the options,$120^{\circ}$ is the intended answer.

(A) We know that the opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the center of the circle.
Therefore,$\angle AOB + \angle COD = 180^{\circ}$.
Given that $\angle AOB = 125^{\circ}$,we have:
$\angle COD = 180^{\circ} - 125^{\circ}$
$\angle COD = 55^{\circ}$.

(B) In the figure,$AOC$ is a diameter of the circle. We know that the angle subtended by a diameter at any point on the circle is $90^{\circ}$.
Therefore,$\angle ABC = 90^{\circ}$.
In $\triangle ACB$,the sum of all interior angles is $180^{\circ}$.
So,$\angle BAC + \angle ABC + \angle ACB = 180^{\circ}$.
$\angle BAC + 90^{\circ} + 50^{\circ} = 180^{\circ}$.
$\angle BAC + 140^{\circ} = 180^{\circ}$.
$\angle BAC = 180^{\circ} - 140^{\circ} = 40^{\circ}$.
Since $AT$ is the tangent to the circle at point $A$,the radius $OA$ is perpendicular to the tangent $AT$.
Therefore,$\angle OAT = 90^{\circ}$.
Since $A, O, C$ are collinear,$\angle OAT$ is the same as $\angle CAT = 90^{\circ}$.
We can write $\angle CAT = \angle CAB + \angle BAT$.
$90^{\circ} = 40^{\circ} + \angle BAT$.
$\angle BAT = 90^{\circ} - 40^{\circ} = 50^{\circ}$.
Thus,the value of $\angle BAT$ is $50^{\circ}$.

(C) Given: Radius of the circle $OQ = OR = 5 \, cm$. Distance from centre $O$ to point $P$ is $OP = 13 \, cm$.
Since $PQ$ and $PR$ are tangents,the radius is perpendicular to the tangent at the point of contact. Therefore,$\angle OQP = 90^\circ$ and $\angle ORP = 90^\circ$.
In the right-angled $\triangle OQP$,by the Pythagorean theorem:
$OP^2 = OQ^2 + PQ^2$
$13^2 = 5^2 + PQ^2$
$169 = 25 + PQ^2$
$PQ^2 = 169 - 25 = 144$
$PQ = 12 \, cm$.
Now,the area of $\triangle OQP = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times OQ \times PQ = \frac{1}{2} \times 5 \times 12 = 30 \, cm^2$.
The quadrilateral $PQOR$ is composed of two congruent right-angled triangles,$\triangle OQP$ and $\triangle ORP$.
Therefore,the area of quadrilateral $PQOR = 2 \times \text{Area}(\triangle OQP) = 2 \times 30 = 60 \, cm^2$.

(D) Let the circle have center $O$ and radius $r = 5 \, cm$. $AB$ is a diameter,so $OA = OB = 5 \, cm$.
Tangent $XY$ is drawn at point $A$. Since the tangent at any point of a circle is perpendicular to the radius through the point of contact,$OA \perp XY$.
Let the chord $CD$ be parallel to $XY$. Let $CD$ intersect the diameter $AB$ at point $E$. Since $CD \parallel XY$ and $OA \perp XY$,it follows that $OE \perp CD$.
The distance of the chord $CD$ from $A$ is given as $AE = 8 \, cm$.
Since $OA = 5 \, cm$,the distance of the chord from the center $O$ is $OE = AE - OA = 8 - 5 = 3 \, cm$.
In the right-angled triangle $\triangle OEC$,by the Pythagorean theorem:
$OC^2 = OE^2 + EC^2$
$5^2 = 3^2 + EC^2$
$25 = 9 + EC^2$
$EC^2 = 16$
$EC = 4 \, cm$.
Since the perpendicular from the center to a chord bisects the chord,$CD = 2 \times EC = 2 \times 4 = 8 \, cm$.

(B) Given,$\angle QPR = 50^{\circ}$.
We know that the tangent at any point of a circle is perpendicular to the radius through the point of contact.
Therefore,$\angle OPR = 90^{\circ}$.
From the figure,$\angle OPQ + \angle QPR = \angle OPR = 90^{\circ}$.
Substituting the given value,$\angle OPQ + 50^{\circ} = 90^{\circ}$.
Thus,$\angle OPQ = 90^{\circ} - 50^{\circ} = 40^{\circ}$.
In $\triangle OPQ$,$OP = OQ$ (radii of the same circle).
Since angles opposite to equal sides are equal,$\angle OQP = \angle OPQ = 40^{\circ}$.
Using the angle sum property of a triangle in $\triangle OPQ$:
$\angle POQ + \angle OPQ + \angle OQP = 180^{\circ}$.
$\angle POQ + 40^{\circ} + 40^{\circ} = 180^{\circ}$.
$\angle POQ + 80^{\circ} = 180^{\circ}$.
$\angle POQ = 180^{\circ} - 80^{\circ} = 100^{\circ}$.

(A) Given that $AB \parallel PR$ and $PQR$ is a tangent at $Q$.
By the alternate segment theorem,the angle between the tangent $QR$ and the chord $BQ$ is equal to the angle subtended by the chord in the alternate segment.
Therefore,$\angle BAQ = \angle BQR = 70^{\circ}$.
Since $AB \parallel PR$,the alternate interior angles are equal.
Thus,$\angle ABQ = \angle BQR = 70^{\circ}$.
In $\triangle ABQ$,the sum of angles is $180^{\circ}$.
$\angle AQB + \angle ABQ + \angle BAQ = 180^{\circ}$.
$\angle AQB + 70^{\circ} + 70^{\circ} = 180^{\circ}$.
$\angle AQB + 140^{\circ} = 180^{\circ}$.
$\angle AQB = 180^{\circ} - 140^{\circ} = 40^{\circ}$.

(A) True.
In $\triangle PBO$,$OB = OP$ (radii of the same circle).
Therefore,$\angle OPB = \angle PBO = 30^{\circ}$.
By angle sum property in $\triangle PBO$,$\angle POB = 180^{\circ} - (30^{\circ} + 30^{\circ}) = 120^{\circ}$.
Since $BOA$ is a straight line,$\angle POA = 180^{\circ} - 120^{\circ} = 60^{\circ}$.
In $\triangle OPT$,$OP \perp PT$ (radius is perpendicular to the tangent at the point of contact).
Therefore,$\angle OPT = 90^{\circ}$.
In $\triangle OPT$,$\angle PTA = 180^{\circ} - (\angle POT + \angle OPT) = 180^{\circ} - (60^{\circ} + 90^{\circ}) = 180^{\circ} - 150^{\circ} = 30^{\circ}$.
Thus,the statement is True.

(B) False.
Since $QL$ is a tangent at $Q$,the angle between the tangent and the chord $SQ$ is equal to the angle in the alternate segment. However,using the property of the radius being perpendicular to the tangent: $\angle OQS = 90^{\circ} - \angle SQL = 90^{\circ} - 50^{\circ} = 40^{\circ}$.
In $\triangle OSQ$,$OS = OQ$ (radii of the same circle),so $\angle OSQ = \angle OQS = 40^{\circ}$.
Similarly,for the tangent $RM$ at $R$,$\angle ORS = 90^{\circ} - \angle SRM = 90^{\circ} - 60^{\circ} = 30^{\circ}$.
In $\triangle OSR$,$OS = OR$ (radii of the same circle),so $\angle OSR = \angle ORS = 30^{\circ}$.
Therefore,$\angle QSR = \angle OSQ + \angle OSR = 40^{\circ} + 30^{\circ} = 70^{\circ}$.

(TRUE) True.
Let there be a number of circles passing through the end points $P$ and $Q$ of a line segment $PQ$.
Since $PQ$ is a common chord for all these circles,the centre of any circle passing through $P$ and $Q$ must be equidistant from $P$ and $Q$.
We know that the locus of points equidistant from two fixed points $P$ and $Q$ is the perpendicular bisector of the line segment $PQ$.
Therefore,the centres of all such circles must lie on the perpendicular bisector of $PQ$.

(N/A) Given: Two tangents $PQ$ and $PR$ are drawn from an external point $P$ to a circle with centre $O$.
To prove: $QORP$ is a cyclic quadrilateral.
Proof:
$1$. Since $PQ$ and $PR$ are tangents to the circle at points $Q$ and $R$ respectively,the radius is perpendicular to the tangent at the point of contact.
$2$. Therefore,$OQ \perp PQ$ and $OR \perp PR$.
$3$. This implies $\angle OQP = 90^{\circ}$ and $\angle ORP = 90^{\circ}$.
$4$. In quadrilateral $QORP$,the sum of the opposite angles is $\angle OQP + \angle ORP = 90^{\circ} + 90^{\circ} = 180^{\circ}$.
$5$. Since the sum of a pair of opposite angles in the quadrilateral $QORP$ is $180^{\circ}$,it is a cyclic quadrilateral.
Hence proved.

(N/A) Given: $AB$ and $CD$ are common tangents to two circles of unequal radii.
To prove: $AB = CD$.
Construction: Produce $AB$ and $CD$ to intersect at point $P$.
Proof:
Since $PA$ and $PC$ are tangents drawn from an external point $P$ to the larger circle,their lengths are equal:
$PA = PC$ ... $(1)$
Since $PB$ and $PD$ are tangents drawn from the same external point $P$ to the smaller circle,their lengths are equal:
$PB = PD$ ... $(2)$
Subtracting equation $(2)$ from equation $(1)$:
$PA - PB = PC - PD$
$AB = CD$
Hence proved.

(N/A) Let $PR$ be a chord of a circle. Tangents are drawn at the points $P$ and $R$. Let these tangents meet at a point $T$ outside the circle.
In $\triangle TPR$,$TP = TR$ (tangents drawn from an external point to a circle are equal in length).
Since $TP = TR$,the angles opposite to these sides are equal,i.e.,$\angle TPR = \angle TRP$.
These angles are the angles made by the tangents with the chord $PR$.
Hence,the tangents drawn at the ends of a chord of a circle make equal angles with the chord.

(N/A) Given: $AB$ is a diameter of the circle with center $O$.
Let $MAN$ be the tangent to the circle at point $A$.
Let $CD$ be any chord of the circle that is parallel to the tangent $MAN$.
Since $OA$ is the radius at the point of contact $A$,we have $OA \perp MAN$.
Since $CD \parallel MAN$ and $OA \perp MAN$,it follows that $OA \perp CD$.
Let the diameter $AB$ intersect the chord $CD$ at point $E$.
Thus,$OE \perp CD$.
According to the theorem,the perpendicular drawn from the center of a circle to a chord bisects the chord.
Therefore,$CE = ED$.
Since $CD$ was an arbitrary chord parallel to the tangent at $A$,the diameter $AB$ bisects all such chords.

(N/A) $(i)$ We know that the perpendicular from the center of a circle to a chord bisects the chord. Since $ON \perp AB$,we have $AN = BN$.
Now,$PA \cdot PB = (PN - AN)(PN + BN)$.
Since $AN = BN$,we can write:
$PA \cdot PB = (PN - AN)(PN + AN) = PN^2 - AN^2$.
$(ii)$ In the right-angled triangle $\triangle ONP$,by the Pythagorean theorem:
$OP^2 = ON^2 + PN^2 \implies PN^2 = OP^2 - ON^2$.
Substituting this into the expression $PN^2 - AN^2$:
$PN^2 - AN^2 = (OP^2 - ON^2) - AN^2 = OP^2 - (ON^2 + AN^2)$.
In the right-angled triangle $\triangle ONA$,by the Pythagorean theorem,$ON^2 + AN^2 = OA^2$.
So,$PN^2 - AN^2 = OP^2 - OA^2$.
Since $OA$ and $OT$ are both radii of the same circle,$OA = OT$.
Therefore,$PN^2 - AN^2 = OP^2 - OT^2$.
$(iii)$ From $(i)$ and $(ii)$,we have $PA \cdot PB = OP^2 - OT^2$.
In the right-angled triangle $\triangle OTP$ (where $\angle OTP = 90^{\circ}$ because the tangent is perpendicular to the radius at the point of contact),by the Pythagorean theorem:
$OP^2 = OT^2 + PT^2 \implies OP^2 - OT^2 = PT^2$.
Thus,$PA \cdot PB = PT^2$.

(N/A) As shown in the figure,the circle touches the sides of the triangle $ABC$ at points $P$,$Q$,and $R$.
According to the theorem,the lengths of tangents drawn from an external point to a circle are equal.
Therefore,we have:
$BQ = BP$
$CP = CR$
$AQ = AR$
Now,consider the perimeter of the triangle $ABC$:
$Perimeter = AB + BC + AC$
$Perimeter = AB + (BP + CP) + AC$
Substituting the values from the tangent theorem:
$Perimeter = AB + BQ + CR + AC$
$Perimeter = (AB + BQ) + (AC + CR)$
$Perimeter = AQ + AR$
Since $AQ = AR$,we can write:
$Perimeter = AQ + AQ = 2AQ$
Thus,$2AQ = AB + BC + AC$
$AQ = \frac{1}{2}(AB + BC + AC)$

(N/A) Let the points of contact of the circle with the sides $AB, BC, CD, DE, EF,$ and $FA$ be $P, Q, R, S, T,$ and $U$ respectively.
Since the lengths of tangents drawn from an external point to a circle are equal,we have:
$AP = AU$
$BP = BQ$
$CQ = CR$
$DR = DS$
$ES = ET$
$FT = FU$
Now,consider the sum of alternate sides:
$AB + CD + EF = (AP + PB) + (CR + RD) + (ET + TF)$
Substituting the tangent equalities:
$AB + CD + EF = (AU + BQ) + (CQ + DS) + (ES + FU)$
Rearranging the terms:
$AB + CD + EF = (BQ + CQ) + (DS + ES) + (FU + AU)$
$AB + CD + EF = BC + DE + FA$
Hence,it is proved.

(N/A) Since $AC$ is a diameter,the angle in a semi-circle is $90^{\circ}$.
Therefore,$\angle ABC = 90^{\circ}$ (Angle in a semi-circle property).
In $\Delta ABC$,the sum of interior angles is $180^{\circ}$,so $\angle CAB + \angle ABC + \angle ACB = 180^{\circ}$.
Substituting $\angle ABC = 90^{\circ}$,we get $\angle CAB + \angle ACB = 180^{\circ} - 90^{\circ} = 90^{\circ}$ .......$(i)$
Since $AT$ is a tangent at $A$ and $AC$ is the radius (part of the diameter),the tangent is perpendicular to the radius at the point of contact.
Thus,$\angle CAT = 90^{\circ}$.
This implies $\angle CAB + \angle BAT = 90^{\circ}$ ..........$(ii)$
From equations $(i)$ and $(ii)$:
$\angle CAB + \angle ACB = \angle CAB + \angle BAT$
Subtracting $\angle CAB$ from both sides,we get $\angle ACB = \angle BAT$. Hence proved.

(D) Given,two circles have radii $OP = 3\, cm$ and $PO^{\prime} = 4\, cm$.
These two circles intersect at $P$ and $Q$.
Since $OP$ and $O^{\prime}P$ are radii to the points of contact of the tangents at $P$,and the problem states $OP$ and $O^{\prime}P$ are perpendicular to each other at $P$,we have $\angle OPO^{\prime} = 90^{\circ}$.
In right-angled $\triangle OPO^{\prime}$,by Pythagoras theorem:
$(OO^{\prime})^2 = (OP)^2 + (PO^{\prime})^2 = 3^2 + 4^2 = 9 + 16 = 25$.
So,$OO^{\prime} = 5\, cm$.
Let $PN$ be the perpendicular from $P$ to $OO^{\prime}$,where $N$ is the intersection point on $OO^{\prime}$.
Let $ON = x$,then $NO^{\prime} = 5 - x$.
In right-angled $\triangle OPN$,$(PN)^2 = (OP)^2 - (ON)^2 = 3^2 - x^2 = 9 - x^2$ ... $(i)$
In right-angled $\triangle PNO^{\prime}$,$(PN)^2 = (PO^{\prime})^2 - (NO^{\prime})^2 = 4^2 - (5 - x)^2 = 16 - (25 + x^2 - 10x) = 16 - 25 - x^2 + 10x = 10x - 9 - x^2$ ... (ii)
Equating $(i)$ and (ii):
$9 - x^2 = 10x - 9 - x^2$
$18 = 10x \Rightarrow x = 1.8\, cm$.
Substituting $x = 1.8$ in $(i)$:
$(PN)^2 = 9 - (1.8)^2 = 9 - 3.24 = 5.76$.
$PN = \sqrt{5.76} = 2.4\, cm$.
The common chord $PQ = 2 \times PN = 2 \times 2.4 = 4.8\, cm$.

(B) $PQ$ and $PR$ are two tangents drawn from an external point $P$.
$\therefore PQ = PR$ [The lengths of tangents drawn from an external point to a circle are equal].
In $\triangle PQR$,since $PQ = PR$,the angles opposite to equal sides are equal.
$\therefore \angle PQR = \angle QRP$.
In $\triangle PQR$,the sum of angles is $180^{\circ}$.
$\angle PQR + \angle QRP + \angle RPQ = 180^{\circ}$
$2 \angle PQR + 30^{\circ} = 180^{\circ}$
$2 \angle PQR = 150^{\circ}$
$\angle PQR = 75^{\circ}$.
Since chord $SR \parallel$ tangent $PQ$,the alternate interior angles are equal.
$\therefore \angle SRQ = \angle RQP = 75^{\circ}$.
By the alternate segment theorem,the angle between the tangent $PR$ and chord $RQ$ is equal to the angle subtended by the chord in the alternate segment.
$\therefore \angle PQR = \angle QSR = 75^{\circ}$.
In $\triangle QRS$,the sum of angles is $180^{\circ}$.
$\angle RQS + \angle QSR + \angle SRQ = 180^{\circ}$
$\angle RQS + 75^{\circ} + 75^{\circ} = 180^{\circ}$
$\angle RQS + 150^{\circ} = 180^{\circ}$
$\angle RQS = 30^{\circ}$.

(N/A) Let $M$ be the mid-point of an arc $AMB$ and $TMT'$ be the tangent to the circle at $M$.
Join $AB$,$AM$,and $MB$.
Since,$\text{arc } AM = \text{arc } MB$,
$\Rightarrow$ Chord $AM =$ Chord $MB$.
In $\triangle AMB$,$AM = MB$.
$\Rightarrow \angle MAB = \angle MBA$ (angles opposite to equal sides are equal) $\dots(i)$.
Since $TMT'$ is a tangent line at $M$,by the alternate segment theorem:
$\angle AMT = \angle MBA$ (angles in alternate segments are equal).
From $(i)$,$\angle AMT = \angle MAB$.
But $\angle AMT$ and $\angle MAB$ are alternate interior angles formed by the transversal $AM$ intersecting lines $AB$ and $TMT'$.
Since the alternate interior angles are equal,the lines must be parallel.
Therefore,$AB \parallel TMT'$.
Hence,the tangent drawn at the mid-point of an arc of a circle is parallel to the chord joining the end points of the arc.

(B) Given,$OT = 13 \, cm$ and radius $OP = 5 \, cm$.
Since the tangent at any point of a circle is perpendicular to the radius through the point of contact,$OP \perp PT$.
In right-angled $\triangle OPT$,by Pythagoras theorem:
$PT^2 = OT^2 - OP^2 = 13^2 - 5^2 = 169 - 25 = 144$.
So,$PT = 12 \, cm$.
Since the lengths of tangents drawn from an external point to a circle are equal,$PT = QT = 12 \, cm$.
Also,$ET = OT - OE = 13 - 5 = 8 \, cm$.
Let $PA = AE = x$. Then $AT = PT - PA = 12 - x$.
In right-angled $\triangle AET$,by Pythagoras theorem:
$AT^2 = AE^2 + ET^2$
$(12 - x)^2 = x^2 + 8^2$
$144 - 24x + x^2 = x^2 + 64$
$24x = 144 - 64 = 80$
$x = \frac{80}{24} = \frac{10}{3} \, cm$.
Similarly,$EB = QB = \frac{10}{3} \, cm$.
Therefore,$AB = AE + EB = \frac{10}{3} + \frac{10}{3} = \frac{20}{3} \, cm$.

(D) Let $O$ be the center of the circle with radius $R = 9\, cm$. Let $AM$ be the altitude from $A$ to $BC$. Since $\triangle ABC$ is isosceles with $AB = AC$, the altitude $AM$ passes through the center $O$.
Let $AM = x$. Then $OM = |x - 9|$.
In $\triangle AMC$, by Pythagoras theorem: $MC^2 = AC^2 - AM^2 = 6^2 - x^2 = 36 - x^2$.
In $\triangle OMC$, by Pythagoras theorem: $MC^2 = OC^2 - OM^2 = 9^2 - (x - 9)^2 = 81 - (x^2 - 18x + 81) = 18x - x^2$.
Equating the two expressions for $MC^2$: $36 - x^2 = 18x - x^2$.
$18x = 36 \Rightarrow x = 2\, cm$.
Thus, $AM = 2\, cm$.
Now, $MC^2 = 36 - 2^2 = 36 - 4 = 32$.
$MC = \sqrt{32} = 4\sqrt{2}\, cm$.
Since $AM$ bisects $BC$, $BC = 2 \times MC = 8\sqrt{2}\, cm$.
Area of $\triangle ABC = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times BC \times AM = \frac{1}{2} \times 8\sqrt{2} \times 2 = 8\sqrt{2}\, cm^{2}$.

(B) Let the radius of the incircle be $r$.
Given,$AB = 8$,$BC = 15$,and $\angle B = 90^{\circ}$.
Using the Pythagoras theorem in $\Delta ABC$:
$AC^{2} = AB^{2} + BC^{2}$
$AC^{2} = 8^{2} + 15^{2}$
$AC^{2} = 64 + 225 = 289$
$AC = \sqrt{289} = 17$.
The area of $\Delta ABC$ can be expressed as the sum of the areas of $\Delta AOB$,$\Delta BOC$,and $\Delta AOC$,where $O$ is the incenter:
Area of $\Delta ABC = \text{Area}(\Delta AOB) + \text{Area}(\Delta BOC) + \text{Area}(\Delta AOC)$
$\frac{1}{2} \times AB \times BC = \frac{1}{2} \times AB \times r + \frac{1}{2} \times BC \times r + \frac{1}{2} \times AC \times r$
$\frac{1}{2} \times 8 \times 15 = \frac{1}{2} \times r \times (AB + BC + AC)$
$60 = \frac{1}{2} \times r \times (8 + 15 + 17)$
$60 = \frac{1}{2} \times r \times 40$
$60 = 20r$
$r = \frac{60}{20} = 3$.
Alternatively,for a right-angled triangle,the inradius $r = \frac{AB + BC - AC}{2} = \frac{8 + 15 - 17}{2} = \frac{6}{2} = 3$.

(D) Let $R$ be the intersection of $OP$ and $AB$. Since $PA$ and $PB$ are tangents from an external point $P$,$PA = PB$ and $\triangle OAP \cong \triangle OBP$. Thus,$OP$ is the perpendicular bisector of chord $AB$.
Given $AB = 24$,so $AR = RB = 12$.
In right-angled $\triangle ORA$,by Pythagoras theorem:
$OR^2 = OA^2 - AR^2 = 13^2 - 12^2 = 169 - 144 = 25$.
So,$OR = 5$.
In $\triangle OAP$,$\angle OAP = 90^\circ$ (tangent is perpendicular to radius at point of contact). $AR$ is the altitude to the hypotenuse $OP$.
By property of right triangles,$OA^2 = OR \cdot OP$.
$13^2 = 5 \cdot OP \implies 169 = 5 \cdot OP \implies OP = \frac{169}{5} = 33.8$.
In right-angled $\triangle OAP$,$PA^2 = OP^2 - OA^2 = (33.8)^2 - 13^2 = 1142.44 - 169 = 973.44$.
$PA = \sqrt{973.44} = 31.2 = \frac{156}{5}$.

(D) In $\odot(O, 15)$,$OA = OB = 15$ (radius) and $AB = 30$ (diameter).
In $\odot(O, 9)$,$OD = 9$ (radius).
Since $BD$ is a tangent to $\odot(O, 9)$ at $D$,we have $\overline{OD} \perp \overline{BD}$. Thus,$\triangle ODB$ is a right-angled triangle with $\angle ODB = 90^{\circ}$.
Since $AB$ is a diameter of $\odot(O, 15)$,any angle inscribed in a semicircle is a right angle. Therefore,$\angle ACB = 90^{\circ}$.
Now,consider $\triangle ODB$ and $\triangle ACB$:
$1$. $\angle ODB = \angle ACB = 90^{\circ}$
$2$. $\angle DBO = \angle CBA$ (Common angle)
By $AA$ similarity criterion,$\triangle ODB \sim \triangle ACB$.
Therefore,the ratios of corresponding sides are equal:
$\frac{AC}{OD} = \frac{AB}{OB}$
$\frac{AC}{9} = \frac{30}{15}$
$\frac{AC}{9} = 2$
$AC = 9 \times 2 = 18$.

(A) In a cyclic quadrilateral,the sum of the measures of opposite angles is $180^{\circ}$.
Given that $ABCD$ is a cyclic quadrilateral,$\angle B$ and $\angle D$ are opposite angles.
Therefore,$m \angle B + m \angle D = 180^{\circ}$.
Substituting the given value,$60^{\circ} + m \angle D = 180^{\circ}$.
Thus,$m \angle D = 180^{\circ} - 60^{\circ} = 120^{\circ}$.

(D) From the figure,$PA$ and $PB$ are tangents to the circle from an external point $P$.
According to the theorem,the lengths of tangents drawn from an external point to a circle are equal,so $PA = PB$.
In $\triangle PAB$,since $PA = PB$,it is an isosceles triangle.
Therefore,the angles opposite to equal sides are equal,which means $m \angle PBA = m \angle PAB$.
From the figure,the angle between the tangent $PA$ and the chord $AB$ is given as $55^{\circ}$.
However,the angle $m \angle PAB$ is the angle between the tangent $PA$ and the chord $AB$.
Thus,$m \angle PAB = 55^{\circ}$.
Since $m \angle PBA = m \angle PAB$,we have $m \angle PBA = 55^{\circ}$.

(C) According to the theorem,the lengths of tangents drawn from an external point to a circle are equal.
From point $A$,$AB$ and $AC$ are tangents to the circle,so $AB = AC = 3$.
From point $P$,$PB$ and $PS$ are tangents to the circle,so $PB = PS$.
From point $Q$,$QC$ and $QS$ are tangents to the circle,so $QC = QS$.
The perimeter of $\Delta APQ = AP + PQ + AQ$.
Since $PQ = PS + SQ$,we can substitute $PB$ for $PS$ and $QC$ for $SQ$.
Perimeter $= AP + (PB + QC) + AQ$.
Perimeter $= (AP + PB) + (AQ + QC) = AB + AC$.
Since $AB = 3$ and $AC = 3$,the perimeter $= 3 + 3 = 6$.

(A) Let the radius of the circle be $r$. The circle is the incircle of the right-angled triangle $ABC$,where $AB = 3$ and $BC = 4$.
Using the Pythagorean theorem,the hypotenuse $AC = \sqrt{AB^2 + BC^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.
The radius $r$ of the incircle of a right-angled triangle is given by the formula $r = \frac{AB + BC - AC}{2}$.
Substituting the values,$r = \frac{3 + 4 - 5}{2} = \frac{2}{2} = 1$.
Thus,the radius of the circle is $1$.

(B) The diameter of the semicircle is $d = 50 \, cm$,so the radius is $r = 25 \, cm$.
For a triangle inscribed in a semicircle,the base of the triangle is the diameter of the semicircle $(b = 50 \, cm)$.
The height of the triangle is the radius of the semicircle $(h = r = 25 \, cm)$ because the maximum height is achieved when the vertex is at the midpoint of the arc.
The area of the triangle is given by the formula $A = \frac{1}{2} \times \text{base} \times \text{height}$.
Substituting the values: $A = \frac{1}{2} \times 50 \times 25 = 25 \times 25 = 625 \, cm^{2}$.

(C) Let the lengths of the tangents from vertices $A, B, C$ to the circle be $x, y, z$ respectively.
Since tangents drawn from an external point to a circle are equal in length,we have:
$AD = AF = x$
$BD = BE = y$
$CE = CF = z$
Given the side lengths of $\Delta ABC$:
$AB = x + y = 13$
$BC = y + z = 12$
$CA = z + x = 5$
Adding these three equations:
$(x + y) + (y + z) + (z + x) = 13 + 12 + 5$
$2(x + y + z) = 30$
$x + y + z = 15$
We need to find $AD = x$. Using $y + z = 12$:
$x + (y + z) = 15$
$x + 12 = 15$
$x = 15 - 12 = 3$
Therefore,$AD = 3$.

(D) The circle $\odot(O, 5)$ has a radius $r = 5$.
Since the circle touches all sides of the square,the diameter of the circle is equal to the side length of the square.
Side length of the square $= 2 \times r = 2 \times 5 = 10$.
The perimeter of a square $= 4 \times \text{side length} = 4 \times 10 = 40$.

(B) circle touches all the four sides of $\square ABCD$. According to the property of a tangential quadrilateral,the sums of opposite sides are equal.
$\therefore AB + CD = AD + BC$
Substituting the given values:
$\therefore 5 + 6 = AD + 8$
$\therefore 11 = AD + 8$
$\therefore AD = 11 - 8 = 3$

(C) $\square PQRS$ is a cyclic quadrilateral.
In a cyclic quadrilateral, the sum of the measures of opposite angles is $180^{\circ}$.
Therefore, $m \angle P + m \angle R = 180^{\circ}$.
Given that $m \angle P = 30^{\circ}$.
Substituting the value, we get $30^{\circ} + m \angle R = 180^{\circ}$.
Thus, $m \angle R = 180^{\circ} - 30^{\circ} = 150^{\circ}$.

(A) Let the tangent $\overleftrightarrow{PQ}$ touch the circle at point $R$.
Since tangents drawn from an external point to a circle are equal in length,we have:
$AB = AC$ (tangents from point $A$)
$PB = PR$ (tangents from point $P$)
$QC = QR$ (tangents from point $Q$)
The perimeter of $\Delta APQ = AP + PQ + AQ$.
Substituting $PQ = PR + QR$,we get:
Perimeter $= AP + (PR + QR) + AQ$.
Using the equalities $PR = PB$ and $QR = QC$:
Perimeter $= AP + PB + QC + AQ$.
Since $AP + PB = AB$ and $AQ + QC = AC$:
Perimeter $= AB + AC$.
Since $AB = AC$,the perimeter $= AB + AB = 2 AB$ (or $2 AC$).
Thus,the correct option is $2 AB$.

(A) For a quadrilateral circumscribing a circle,the sum of opposite sides is equal,i.e.,$AB + CD = BC + AD$.
Given that $\overline{AB}$ is the largest side,let $AB = x$.
Since the circle is tangent to all sides,the lengths of the tangents from the vertices to the circle are equal. Let the lengths of the tangents from vertices $A, B, C, D$ be $a, b, c, d$ respectively.
Then $AB = a + b$,$BC = b + c$,$CD = c + d$,and $AD = a + d$.
Since $AB$ is the largest side,$a+b$ is the maximum value.
In a tangential quadrilateral,the side opposite to the largest side is generally the smallest side. Thus,$\overline{CD}$ is the smallest side.

(C) Let the tangents from point $A$ to the circle be $AB$ and $AC$. By the theorem,lengths of tangents drawn from an external point to a circle are equal,so $AB = AC$.
Let the tangent $PQ$ touch the circle at point $M$. Then $PM = PB$ and $QM = QC$ (tangents from points $P$ and $Q$ respectively).
The perimeter of $\Delta APQ = AP + PQ + AQ = AP + (PM + MQ) + AQ$.
Substituting the equal tangent lengths: Perimeter $= AP + PB + QC + AQ = (AP + PB) + (AQ + QC) = AB + AC$.
Since $AB = AC$,the perimeter $= 2 AB$.
Given that the perimeter is $16$,we have $2 AB = 16$,which implies $AB = 8$.

(C) The distance from the centre $P$ of the circle to the line $AB$ is the length of the perpendicular segment $PQ$.
Given that $Q$ lies in the interior of the circle,the length of $PQ$ must be less than the radius $r$ of the circle (i.e.,$PQ < r$).
According to the properties of a circle,if the perpendicular distance from the centre to a line is less than the radius of the circle,then the line is a secant line and intersects the circle at two distinct points.

(B) In $\Delta ABC$,$m\angle B = 90^{\circ}$,$AB = 24$,and $BC = 7$.
Using the Pythagoras theorem:
$AC^2 = AB^2 + BC^2$
$AC^2 = (24)^2 + (7)^2$
$AC^2 = 576 + 49$
$AC^2 = 625$
$AC = \sqrt{625} = 25$.
The radius $r$ of the incircle of a right-angled triangle is given by the formula:
$r = \frac{AB + BC - AC}{2}$
$r = \frac{24 + 7 - 25}{2}$
$r = \frac{31 - 25}{2}$
$r = \frac{6}{2} = 3$.
Therefore,the radius of the circle is $3$.