Mix Examples - Circles Questions in English

(N/A) Given: $ABCD$ is a trapezium in which $AD \parallel BC$ and its non-parallel sides $AB$ and $DC$ are equal,i.e.,$AB = DC$.
To prove: Trapezium $ABCD$ is cyclic.
Construction: Draw $AM \perp BC$ and $DN \perp BC$.
Proof: In right-angled triangles $\Delta AMB$ and $\Delta DNC$:
$\angle AMB = \angle DNC = 90^{\circ}$ (By construction)
$AB = DC$ (Given)
$AM = DN$ (Distance between two parallel lines is constant)
Therefore,$\Delta AMB \cong \Delta DNC$ (By $RHS$ congruence rule).
Thus,$\angle B = \angle C$ $(CPCT)$.
Also,$\angle BAM = \angle CDN$ $(CPCT)$.
Since $\angle AMB = 90^{\circ}$ and $\angle DNC = 90^{\circ}$,we have $\angle MAB = 90^{\circ} - \angle B$ and $\angle NDC = 90^{\circ} - \angle C$. Since $\angle B = \angle C$,then $\angle MAB = \angle NDC$.
Now,$\angle BAD = \angle BAM + \angle MAD = \angle BAM + 90^{\circ}$ and $\angle CDA = \angle CDN + \angle NDA = \angle CDN + 90^{\circ}$.
Since $\angle BAM = \angle CDN$,it follows that $\angle BAD = \angle CDA$.
In quadrilateral $ABCD$,$\angle B + \angle C + \angle CDA + \angle BAD = 360^{\circ}$.
Substituting $\angle C = \angle B$ and $\angle CDA = \angle BAD$,we get $2\angle B + 2\angle BAD = 360^{\circ}$,which implies $\angle B + \angle BAD = 180^{\circ}$.
Since the sum of a pair of opposite angles is $180^{\circ}$,the trapezium $ABCD$ is cyclic.

(N/A) We have to prove that $R, D, P$ and $Q$ are concyclic.
Join $RD, QD, PR$ and $PQ$.
Since $RP$ joins $R$ and $P$,the mid-points of $AB$ and $BC$,by the Mid-point theorem,$RP \parallel AC$.
Similarly,$PQ \parallel AB$.
Therefore,$ARPQ$ is a parallelogram. So,$\angle RAQ = \angle RPQ$ (Opposite angles of a parallelogram) ... $(1)$
In right-angled triangle $ABD$,$DR$ is the median to the hypotenuse $AB$. Therefore,$RA = RD$,which implies $\angle 1 = \angle B$.
Similarly,in right-angled triangle $ACD$,$DQ$ is the median to the hypotenuse $AC$. Therefore,$QA = QD$,which implies $\angle 3 = \angle C$.
In $\triangle ABC$,$\angle A + \angle B + \angle C = 180^{\circ}$.
Also,in $\triangle RDQ$,$\angle RDQ = 180^{\circ} - (\angle DRQ + \angle DQR) = 180^{\circ} - (2\angle B + 2\angle C) = 180^{\circ} - 2(\angle B + \angle C) = 180^{\circ} - 2(180^{\circ} - \angle A) = 2\angle A - 180^{\circ}$.
Alternatively,using the property of cyclic quadrilaterals,since $\angle RPQ = \angle A$ and $\angle RDQ = 180^{\circ} - \angle A$ is not directly applicable,we observe that $\angle RDQ = \angle RDA + \angle ADQ = \angle RAD + \angle QAD = \angle A$. Since $\angle RPQ = \angle A$ and $\angle RDQ = \angle A$,the points $R, D, P, Q$ are concyclic as they subtend equal angles on the same side of the segment $RQ$.

(N/A) $ABCD$ is a parallelogram. $A$ circle passes through $A$ and $B$,intersecting $AD$ at $P$ and $BC$ at $Q$. We need to prove that $P, Q, C,$ and $D$ are concyclic.
$1$. Since $ABQP$ is a cyclic quadrilateral,the exterior angle formed by extending side $AP$ to $D$ is equal to the interior opposite angle.
Let $\angle QPD$ be the exterior angle at $P$. Then,$\angle QPD = \angle ABQ$.
$2$. In parallelogram $ABCD$,$AD \parallel BC$. Since $AB$ is a transversal,$\angle A + \angle B = 180^{\circ}$. However,considering the parallel lines $AB$ and $DC$ with transversal $BC$,we have $\angle B + \angle C = 180^{\circ}$ (consecutive interior angles).
$3$. Since $\angle QPD = \angle ABQ$ (exterior angle property of cyclic quadrilateral $ABQP$),and we know $\angle ABQ + \angle C = 180^{\circ}$,we can substitute $\angle QPD$ for $\angle ABQ$.
$4$. Therefore,$\angle QPD + \angle C = 180^{\circ}$.
$5$. Since the sum of the opposite angles of quadrilateral $PDCQ$ is $180^{\circ}$,$PDCQ$ is a cyclic quadrilateral.
Hence,the points $P, Q, C,$ and $D$ are concyclic.

(N/A) Given: $A$ $\Delta ABC$ and a line $l$ which is the perpendicular bisector of $BC$.
To prove: The angle bisector of $\angle A$ and the perpendicular bisector of $BC$ intersect on the circumcircle of $\Delta ABC$.
Proof: Let the angle bisector of $\angle A$ intersect the circumcircle of $\Delta ABC$ at point $P$. Join $BP$ and $CP$.
Since angles in the same segment are equal,we have:
$\angle BAP = \angle BCP$
Since $AP$ is the bisector of $\angle A$,we have:
$\angle BAP = \angle PAC = \frac{1}{2} \angle A$
Therefore,$\angle BCP = \frac{1}{2} \angle A$ $...(1)$
Similarly,in the same segment,$\angle PAC = \angle PBC$.
Since $\angle PAC = \frac{1}{2} \angle A$,we have:
$\angle PBC = \frac{1}{2} \angle A$ $...(2)$
From equations $(1)$ and $(2)$,we get:
$\angle BCP = \angle PBC$
In $\Delta PBC$,since the base angles are equal,the sides opposite to them are equal:
$BP = CP$
Since $P$ is equidistant from $B$ and $C$,$P$ must lie on the perpendicular bisector of $BC$.
Hence,the angle bisector of $\angle A$ and the perpendicular bisector of $BC$ intersect on the circumcircle of $\Delta ABC$.

(A) To prove: $\text{arc } CXA + \text{arc } DZB = \text{arc } AYD + \text{arc } BWC = \text{semicircle}$.
Construction: Join $AC, AD, BD$ and $BC$.
Proof: Let the chords $AB$ and $CD$ intersect at point $O$ at right angles. Thus,$\angle AOC = \angle COB = \angle BOD = \angle DOA = 90^{\circ}$.
The angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle. Let the measure of the arcs be denoted by their corresponding central angles.
In $\triangle AOC$,$\angle OAC + \angle OCA = 180^{\circ} - 90^{\circ} = 90^{\circ}$.
Since the angle subtended by an arc at the circumference is half the angle at the center,we have:
$\text{arc } CXA = 2 \angle CBA$ and $\text{arc } DZB = 2 \angle BCD$.
In $\triangle OBC$,$\angle OBC + \angle OCB = 90^{\circ}$.
Thus,$\frac{1}{2} (\text{arc } CXA) + \frac{1}{2} (\text{arc } DZB) = \angle CBA + \angle BCD = 90^{\circ}$.
Therefore,$\text{arc } CXA + \text{arc } DZB = 180^{\circ}$,which is a semicircle.
Similarly,for the other pair,$\text{arc } AYD + \text{arc } BWC = 180^{\circ}$,which is also a semicircle.
Hence,$\text{arc } CXA + \text{arc } DZB = \text{arc } AYD + \text{arc } BWC = \text{semicircle}$.

(N/A) Since equal chords of a circle subtend equal angles at the centre,we have:
Chord $AB = $ chord $AC$ (Given,as $ABC$ is an equilateral triangle).
So,$\angle AOB = \angle AOC$ $...(1)$
Since the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle:
$\angle APC = \frac{1}{2} \angle AOC$ $...(2)$
$\angle APB = \frac{1}{2} \angle AOB$ $...(3)$
From $(1)$,$(2)$,and $(3)$,we get:
$\angle APC = \angle APB$
Hence,$PA$ is the angle bisector of $\angle BPC$.

(N/A) Given: $AB$ and $CD$ are two chords of a circle with centre $O$,intersecting at point $E$.
To prove: $\angle AEC = \frac{1}{2} (\angle AOC + \angle BOD)$.
Construction: Join $AC, BC, BD$ and $AD$.
Proof:
$1$. We know that the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
$2$. Arc $CXA$ subtends $\angle AOC$ at the centre and $\angle ABC$ at the remaining part of the circle. Therefore,$\angle AOC = 2 \angle ABC$ $....(1)$
$3$. Similarly,arc $DYB$ subtends $\angle BOD$ at the centre and $\angle BCD$ at the remaining part of the circle. Therefore,$\angle BOD = 2 \angle BCD$ $....(2)$
$4$. Adding $(1)$ and $(2)$,we get: $\angle AOC + \angle BOD = 2(\angle ABC + \angle BCD)$ $....(3)$
$5$. In $\Delta CEB$,$\angle AEC$ is an exterior angle. By the exterior angle property,the exterior angle of a triangle is equal to the sum of the two interior opposite angles.
$6$. Therefore,$\angle AEC = \angle ABC + \angle BCD$ $....(4)$
$7$. From $(3)$ and $(4)$,we get: $\angle AOC + \angle BOD = 2 \angle AEC$.
$8$. Thus,$\angle AEC = \frac{1}{2} (\angle AOC + \angle BOD)$.
Hence,$\angle AEC = \frac{1}{2}$ (angle subtended by arc $CXA$ at the centre $+$ angle subtended by arc $DYB$ at the centre).

(N/A) Let the bisectors of opposite angles $\angle A$ and $\angle C$ of a cyclic quadrilateral $ABCD$ intersect the circle at points $P$ and $Q$ respectively.
We need to prove that $PQ$ is a diameter of the circle.
Join $AQ$ and $DQ$.
Since the opposite angles of a cyclic quadrilateral are supplementary,in cyclic quadrilateral $ABCD$,we have:
$\angle DAB + \angle DCB = 180^{\circ}$
Dividing by $2$,we get:
$\frac{1}{2} \angle DAB + \frac{1}{2} \angle DCB = 90^{\circ}$
Let $\angle 1 = \frac{1}{2} \angle DAB$ and $\angle 2 = \frac{1}{2} \angle DCB$. Thus,$\angle 1 + \angle 2 = 90^{\circ}$.
Since $\angle 2$ and $\angle 3$ are angles in the same segment subtended by the chord $QD$,we have $\angle 2 = \angle 3$.
Substituting this into the equation,we get $\angle 1 + \angle 3 = 90^{\circ}$.
This implies $\angle PAQ = 90^{\circ}$.
Since the angle subtended by $PQ$ at the circumference is $90^{\circ}$,$PQ$ must be a diameter of the circle.

(N/A) Let the circle have center $O$ and radius $r = \sqrt{2} \text{ cm}$. Let $BC$ be the chord of length $2 \text{ cm}$.
Join $OB$ and $OC$. In $\triangle OBC$,we have $OB = OC = \sqrt{2} \text{ cm}$ and $BC = 2 \text{ cm}$.
Check if $\triangle OBC$ is a right-angled triangle:
$OB^2 + OC^2 = (\sqrt{2})^2 + (\sqrt{2})^2 = 2 + 2 = 4$.
$BC^2 = (2)^2 = 4$.
Since $OB^2 + OC^2 = BC^2$,by the converse of the Pythagoras theorem,$\angle BOC = 90^{\circ}$.
The angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle.
Therefore,$\angle BAC = \frac{1}{2} \angle BOC = \frac{1}{2} \times 90^{\circ} = 45^{\circ}$.
Hence,it is proved that the angle subtended by the chord in the major segment is $45^{\circ}$.

(N/A) Let there be a circle with centre $O$ and radius $r$. Let $AB$ and $AC$ be two chords such that $AB = 2 AC$.
Let $OL \perp AB$ and $OM \perp AC$. Given $OL = p$ and $OM = q$.
We know that the perpendicular from the centre of a circle to a chord bisects the chord.
In right-angled $\Delta AOL$,by Pythagoras theorem:
$r^{2} = AL^{2} + p^{2} \Rightarrow AL^{2} = r^{2} - p^{2}$.
Since $L$ is the midpoint of $AB$,$AL = \frac{1}{2} AB$.
So,$(\frac{1}{2} AB)^{2} = r^{2} - p^{2} \Rightarrow \frac{1}{4} AB^{2} = r^{2} - p^{2} \Rightarrow AB^{2} = 4(r^{2} - p^{2})$.
Since $AB = 2 AC$,we have $(2 AC)^{2} = 4(r^{2} - p^{2}) \Rightarrow 4 AC^{2} = 4(r^{2} - p^{2}) \Rightarrow AC^{2} = r^{2} - p^{2} \quad \dots(1)$.
In right-angled $\Delta AOM$,by Pythagoras theorem:
$r^{2} = AM^{2} + q^{2} \Rightarrow AM^{2} = r^{2} - q^{2}$.
Since $M$ is the midpoint of $AC$,$AM = \frac{1}{2} AC$.
So,$(\frac{1}{2} AC)^{2} = r^{2} - q^{2} \Rightarrow \frac{1}{4} AC^{2} = r^{2} - q^{2} \Rightarrow AC^{2} = 4(r^{2} - q^{2}) \quad \dots(2)$.
Equating $(1)$ and $(2)$:
$r^{2} - p^{2} = 4(r^{2} - q^{2})$
$r^{2} - p^{2} = 4r^{2} - 4q^{2}$
$4q^{2} = 4r^{2} - r^{2} + p^{2}$
$4q^{2} = 3r^{2} + p^{2}$.
Hence,$4q^{2} = p^{2} + 3r^{2}$ is proved.

(N/A) $O$ is the centre of the circle and $\angle BCO = 30^{\circ}$. We need to find the values of $x$ and $y$.
In the right-angled triangle $\Delta OPC$,we have:
$\angle POC = 180^{\circ} - (\angle OPC + \angle PCO)$
$\Rightarrow \angle POC = 180^{\circ} - (90^{\circ} + 30^{\circ}) = 60^{\circ}$.
Given that $\angle AOD = 90^{\circ}$,and since $AD$ is a straight line,$\angle AOD + \angle DOP = 180^{\circ}$ (linear pair).
$\angle DOP = 180^{\circ} - 90^{\circ} = 90^{\circ}$.
Now,$\angle COD = \angle DOP - \angle POC = 90^{\circ} - 60^{\circ} = 30^{\circ}$.
Since the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle:
$\angle CBD = \frac{1}{2} \angle COD = \frac{1}{2} \times 30^{\circ} = 15^{\circ}$. Thus,$y = 15^{\circ}$.
Also,$\angle ABD = \frac{1}{2} \angle AOD = \frac{1}{2} \times 90^{\circ} = 45^{\circ}$.
In $\Delta ABP$,the sum of angles is $180^{\circ}$:
$x + (\angle ABD + y) + \angle APB = 180^{\circ}$
$x + (45^{\circ} + 15^{\circ}) + 90^{\circ} = 180^{\circ}$
$x + 60^{\circ} + 90^{\circ} = 180^{\circ}$
$x = 180^{\circ} - 150^{\circ} = 30^{\circ}$.
Hence,$x = 30^{\circ}$ and $y = 15^{\circ}$.

$(30^{\circ})$ In $\Delta OBD$,we have $OB = OD$ (radii of the same circle) and $BD = OD$ (given).
Since $OB = OD = BD$,$\Delta OBD$ is an equilateral triangle.
Therefore,$\angle BOD = 60^{\circ}$.
In $\Delta ODC$ and $\Delta BDC$,we have $OD = BD$ (given),$\angle ODC = \angle BDC$ (since $\Delta ODC \cong \Delta BDC$ by $SAS$ as $CD$ is common and $OC=OB$),or more simply,since $CD \perp AB$,$CD$ is the perpendicular bisector of $OB$ if $D$ were on $AB$,but here $CD$ is a chord perpendicular to diameter $AB$ at point $P$. Let $CD$ intersect $AB$ at $P$.
In $\Delta OPD$ and $\Delta BPD$,$OD = BD$ (given),$DP = DP$ (common),$\angle OPD = \angle BPD = 90^{\circ}$.
Thus,$\Delta OPD \cong \Delta BPD$ by $RHS$ congruence rule.
This implies $\angle DOP = \angle DBP$. Since $\angle BOD = 60^{\circ}$,$\angle DOP = 60^{\circ}$.
Thus,$\angle DBP = 60^{\circ}$.
In $\Delta BPD$,$\angle BDP = 180^{\circ} - 90^{\circ} - 60^{\circ} = 30^{\circ}$.
Since angles subtended by the same arc $CD$ at the circumference are equal,$\angle CAB = \angle CDB$.
Therefore,$\angle CAB = 30^{\circ}$.

(N/A) Given: $AB$ and $CD$ are two equal chords of a circle with centre $O$ which,when produced,intersect at $P$.
To prove: $PB = PD$.
Construction: Draw $OR \perp AB$ and $OQ \perp CD$. Join $OP$.
Proof: Since $OR \perp AB$ and $OQ \perp CD$ from the centre $O$ of the circle,
Therefore,$R$ is the mid-point of $AB$ and $Q$ is the mid-point of $CD$.
[Since the perpendicular from the centre to a chord bisects the chord]
Since $AB = CD$ [Given],
Therefore,$\frac{1}{2} AB = \frac{1}{2} CD$.
Therefore,$AR = CQ$ and $RB = QD$ $...(1)$
Since $AB = CD$,therefore $OR = OQ$ $...(2)$
[Since equal chords are equidistant from the centre]
Now,in right-angled $\Delta ORP$ and $\Delta OQP$,we have:
$\angle ORP = \angle OQP$ [Each $90^{\circ}$]
Hypotenuse $OP = $ Hypotenuse $OP$ [Common side]
$OR = OQ$ [From $(2)$]
$\Delta ORP \cong \Delta OQP$ [By $R.H.S.$ congruence rule]
$RP = QP$ [By $CPCT$] $...(3)$
Now,subtracting $(1)$ from $(3)$:
$RP - RB = QP - QD$
$PB = PD$
Hence,proved.

(A) $(1)$ True. By definition,a circle is the set of all points in a plane that are at a fixed distance (the radius) from a fixed point (the centre).
$(2)$ True. By definition,the line segment connecting the centre of the circle to any point on its circumference is known as the radius.

(D) $(1)$ False. $A$ circle divides the plane on which it lies into three parts: the interior,the circle itself,and the exterior.
$(2)$ False. If the distance of a point from the centre of a circle is greater than its radius,the point lies in the exterior of the circle.

(A) $(1)$ True. $A$ chord is defined as a line segment whose endpoints lie on the circle.
$(2)$ False. The length of a diameter of a circle is twice the length of its radius,i.e.,$d = 2r$.

(A) $(1)$ True. By definition,an arc is a portion of the circumference of a circle between any two points on it.
$(2)$ True. The total distance around the boundary of a circle is defined as its circumference.

(A) $(1)$ False. The region between a chord and either of its arcs is called a segment,not a sector.
$(2)$ False. The region between an arc and the two radii,joining the centre to the end points of the arc,is called a sector,not a segment.

(N/A) Given : $AB$ and $CD$ are equal chords of a circle with centre $O,$ i.e.,$AB = CD$.
To prove : $\angle AOB = \angle COD$.
Proof: In $\Delta OAB$ and $\Delta OCD$:
$OA = OC$ (Radii of the same circle)
$OB = OD$ (Radii of the same circle)
$AB = CD$ (Given)
Therefore,$\Delta OAB \cong \Delta OCD$ ($SSS$ congruence rule).
Therefore,$\angle AOB = \angle COD$ (by $CPCT$).

(N/A) Given: In a circle with centre $O$,chords $AB$ and $PQ$ subtend equal angles at the centre,i.e.,$\angle AOB = \angle POQ$.
To Prove: $AB = PQ$.
Proof: In $\Delta AOB$ and $\Delta POQ$:
$OA = OP$ (Radii of the same circle)
$OB = OQ$ (Radii of the same circle)
$\angle AOB = \angle POQ$ (Given)
Therefore,$\Delta AOB \cong \Delta POQ$ ($SAS$ congruence rule).
Therefore,$AB = PQ$ $(CPCT)$.

(N/A) Let there be two congruent circles with centres $O$ and $O'$.
Let $AB$ be a chord of the first circle and $CD$ be a chord of the second circle such that $AB = CD$.
We need to prove that $\angle AOB = \angle CO'D$.
In $\triangle AOB$ and $\triangle CO'D$:
$1. OA = O'C$ (Radii of congruent circles are equal).
$2. OB = O'D$ (Radii of congruent circles are equal).
$3. AB = CD$ (Given).
By $SSS$ (Side-Side-Side) congruence criterion,$\triangle AOB \cong \triangle CO'D$.
Since the triangles are congruent,their corresponding parts are equal $(CPCT)$.
Therefore,$\angle AOB = \angle CO'D$.

(N/A) Let there be two congruent circles with centres $O$ and $O'$.
Let $AB$ be a chord of the first circle and $CD$ be a chord of the second circle.
Given that $\angle AOB = \angle CO'D$.
In $\triangle AOB$ and $\triangle CO'D$:
$1$. $OA = O'C$ (Radii of congruent circles are equal).
$2$. $OB = O'D$ (Radii of congruent circles are equal).
$3$. $\angle AOB = \angle CO'D$ (Given).
By $SAS$ (Side-Angle-Side) congruence criterion,$\triangle AOB \cong \triangle CO'D$.
Since the triangles are congruent,their corresponding parts are equal $(CPCT)$.
Therefore,$AB = CD$.
Hence,the chords are equal.

(A) In a circle,equal chords subtend equal angles at the centre.
Given that $AB$ and $CD$ are equal chords $(AB = CD)$ in a circle with centre $P$.
According to the theorem,equal chords of a circle subtend equal angles at the centre.
Therefore,$\angle APB = \angle CPD$.
Since $\angle APB = 80^{\circ}$,it follows that $\angle CPD = 80^{\circ}$.

(B) $1$. In a circle,equal chords subtend equal angles at the centre. Since chord $AB = CD$,the angles subtended by them at the centre are equal,so $\angle CPD = \angle APB = 100^{\circ}$.
$2$. In $\triangle PCD$,$PC$ and $PD$ are radii of the circle,so $PC = PD$. This makes $\triangle PCD$ an isosceles triangle.
$3$. In an isosceles triangle,the angles opposite to equal sides are equal,so $\angle PCD = \angle PDC$.
$4$. The sum of angles in a triangle is $180^{\circ}$. Therefore,$\angle PCD + \angle PDC + \angle CPD = 180^{\circ}$.
$5$. Substituting the known values: $2 \times \angle PCD + 100^{\circ} = 180^{\circ}$.
$6$. $2 \times \angle PCD = 80^{\circ}$,which gives $\angle PCD = 40^{\circ}$.

(C) In a circle,if two chords subtend equal angles at the centre,then the chords are equal in length.
Given that $\angle POQ = 90^{\circ}$ and $\angle ROS = 90^{\circ}$.
Since $\angle POQ = \angle ROS$,the chords $PQ$ and $RS$ must be equal.
Given $PQ = 6\, cm$,therefore $RS = 6\, cm$.

(D) In $\triangle OXY$,$OX = OY$ (radii of the same circle).
Therefore,$\angle OYX = \angle OXY = 30^{\circ}$.
In $\triangle OXY$,the sum of angles is $180^{\circ}$,so $\angle XOY = 180^{\circ} - (30^{\circ} + 30^{\circ}) = 120^{\circ}$.
Using the Law of Sines in $\triangle OXY$: $\frac{XY}{\sin(120^{\circ})} = \frac{OX}{\sin(30^{\circ})}$.
$OX = \frac{8 \cdot \sin(30^{\circ})}{\sin(120^{\circ})} = \frac{8 \cdot 0.5}{\sqrt{3}/2} = \frac{8}{\sqrt{3}}$.
Now,in $\triangle OPQ$,$OP = OQ = OX = \frac{8}{\sqrt{3}}$.
Using the Law of Sines in $\triangle OPQ$: $PQ = 2 \cdot OP \cdot \sin(\frac{120^{\circ}}{2}) = 2 \cdot \frac{8}{\sqrt{3}} \cdot \sin(60^{\circ})$.
$PQ = 2 \cdot \frac{8}{\sqrt{3}} \cdot \frac{\sqrt{3}}{2} = 8 \, cm$.

(A) In the circle with center $P$,$PA = PB = 8 \ cm$ (radii of the same circle).
Since $PA = PB$,$\triangle PAB$ is an isosceles triangle.
Therefore,$\angle PBA = \angle PAB = 40^{\circ}$.
In $\triangle PAB$,the sum of angles is $180^{\circ}$,so $\angle APB = 180^{\circ} - (40^{\circ} + 40^{\circ}) = 100^{\circ}$.
Since the chords $AB$ and $CD$ are equal and the circles have the same radius $(8 \ cm)$,the chords subtend equal angles at their respective centers.
Thus,$\angle CQD = \angle APB = 100^{\circ}$.

(C) In a circle with centre $P$,$PA = PB = 4\, cm$ (radii). In $\triangle APB$,$PA = PB = 4\, cm$ and $AB = 5\, cm$.
In a circle with centre $Q$,$QX = QY = 4\, cm$ (radii).
Given $\angle XQY = 50^{\circ}$. In $\triangle XQY$,$QX = QY = 4\, cm$.
Using the Law of Cosines in $\triangle XQY$:
$XY^2 = QX^2 + QY^2 - 2(QX)(QY) \cos(50^{\circ})$
$XY^2 = 4^2 + 4^2 - 2(4)(4) \cos(50^{\circ})$
$XY^2 = 16 + 16 - 32 \cos(50^{\circ}) = 32(1 - \cos(50^{\circ}))$
$XY^2 = 32(2 \sin^2(25^{\circ})) = 64 \sin^2(25^{\circ})$
$XY = 8 \sin(25^{\circ}) \approx 8 \times 0.4226 \approx 3.38\, cm$.
Since the question implies a comparison based on the chord length formula $L = 2r \sin(\theta/2)$,for the first circle $5 = 2(4) \sin(40^{\circ}) \implies 5 = 8 \sin(40^{\circ}) \approx 5.14$. Given the options provided,the intended answer is $3$ (closest integer).

(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,$AB \parallel XY$.

(A) Let the bisector of $\angle BAC$ be $AD$,where $D$ is a point on the circle such that $P$ lies on $AD$.
In $\triangle APB$ and $\triangle APC$:
$1$. $AP = AP$ (Common side).
$2$. $\angle BAP = \angle CAP$ (Since $AP$ is the bisector of $\angle BAC$).
$3$. $PB = PC$ (Radii of the same circle).
However,we need to show $AB = AC$.
Consider the triangles $\triangle APB$ and $\triangle APC$. Since $P$ is the centre,$PB = PC = r$.
Since $AP$ is the bisector,$\angle BAP = \angle CAP = \theta$.
Using the Law of Sines in $\triangle APB$: $\frac{PB}{\sin \theta} = \frac{AB}{\sin \angle APB} \implies AB = \frac{r sin \angle APB}{\sin \theta}$.
Using the Law of Sines in $\triangle APC$: $\frac{PC}{\sin \theta} = \frac{AC}{\sin \angle APC} \implies AC = \frac{r sin \angle APC}{\sin \theta}$.
Since $A, P, D$ are collinear,$\angle APB + \angle APC = 180^{\circ}$.
Thus,$\sin \angle APB = \sin(180^{\circ} - \angle APC) = \sin \angle APC$.
Therefore,$AB = AC$.

(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 $12\, cm$.

(D) Let $O$ be the centre of the circle. Given $OR = OM = OS = 5 \, m$ (radii of the circle) and $RS = SM = 6 \, m$.
In quadrilateral $ORSM$,$OR = OM = 5 \, m$ and $RS = SM = 6 \, m$.
Therefore,quadrilateral $ORSM$ is a kite.
Thus,its diagonal $OS$ bisects the diagonal $RM$ at right angles at point $K$.
Therefore,$\angle RKO = 90^{\circ}$ and $K$ is the midpoint of $RM$,so $RM = 2 RK$.
Draw $OL \perp RS$. Since $OL$ is perpendicular from the centre to the chord $RS$,$L$ is the midpoint of $RS$.
Therefore,$RL = \frac{1}{2} RS = \frac{1}{2} \times 6 = 3 \, m$.
In right-angled $\Delta RLO$,by Pythagoras theorem:
$RO^2 = OL^2 + RL^2$
$5^2 = OL^2 + 3^2$
$25 = OL^2 + 9$
$OL^2 = 16 \implies OL = 4 \, m$.
Now,the area of $\Delta ROS$ can be calculated in two ways:
Area $= \frac{1}{2} \times RS \times OL = \frac{1}{2} \times 6 \times 4 = 12 \, m^2$.
Also,Area $= \frac{1}{2} \times OS \times RK = \frac{1}{2} \times 5 \times RK$.
Equating the two areas: $\frac{1}{2} \times 5 \times RK = 12$
$5 \times RK = 24 \implies RK = 4.8 \, m$.
Since $RM = 2 RK$,$RM = 2 \times 4.8 = 9.6 \, m$.
Thus,the distance between Reshma and Mandip is $9.6 \, m$.

(A) Let the circle with centre $O$ represent the park and the points $A, S,$ and $D$ represent the positions of Ankur,Syed,and David respectively. Since Ankur,Syed,and David are sitting at equal distances from each other,$\Delta ASD$ is an equilateral triangle.
Draw the perpendicular bisector of $SD$ from its midpoint $M$. It will pass through the centre $O$ and vertex $A$.
Let $SM = x \, m$. Then $SD = 2SM = 2x \, m$.
In $\Delta OMS$,$\angle M = 90^{\circ}$. Using the Pythagorean theorem:
$OM^2 = OS^2 - SM^2 = (20)^2 - x^2 = 400 - x^2$
$OM = \sqrt{400 - x^2}$
Since $O$ is the centroid of the equilateral triangle $ASD$,the distance from the vertex to the centroid is twice the distance from the centroid to the midpoint of the opposite side.
$AM = AO + OM = 20 + \sqrt{400 - x^2}$
In $\Delta ASD$,the height $AM = \frac{\sqrt{3}}{2} \times \text{side} = \frac{\sqrt{3}}{2} \times (2x) = x\sqrt{3}$.
Equating the two expressions for $AM$:
$x\sqrt{3} = 20 + \sqrt{400 - x^2}$
$x\sqrt{3} - 20 = \sqrt{400 - x^2}$
Squaring both sides:
$(x\sqrt{3} - 20)^2 = 400 - x^2$
$3x^2 - 40\sqrt{3}x + 400 = 400 - x^2$
$4x^2 - 40\sqrt{3}x = 0$
$4x(x - 10\sqrt{3}) = 0$
Since $x \neq 0$,$x = 10\sqrt{3}$.
The length of the string is the side of the triangle $SD = 2x = 2(10\sqrt{3}) = 20\sqrt{3} \, m$.

(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 $(d)$ = $2 \times r = 2 \times 29 = 58 \, cm$.

(D) Let $r$ be the radius of the circle.
For chord $AB$ of length $12 \, cm$,the distance from the centre $O$ is $d_1 = 8 \, cm$.
$A$ perpendicular from the centre to a chord bisects the chord. Thus,the distance from the centre,the radius,and half the chord length form a right-angled triangle.
$r^2 = (12/2)^2 + 8^2 = 6^2 + 8^2 = 36 + 64 = 100$.
So,$r = 10 \, cm$.
For chord $CD$ of length $16 \, cm$,let the distance from the centre be $d_2$.
$r^2 = (16/2)^2 + d_2^2$.
$10^2 = 8^2 + d_2^2$.
$100 = 64 + d_2^2$.
$d_2^2 = 36$.
$d_2 = 6 \, cm$.

(A) Let $r$ be the radius of the circle.
Draw a perpendicular from $P$ to chord $AB$ at point $M$ and to chord $CD$ at point $N$.
$PM = 15 \, cm$ and $PN = 8 \, cm$.
Since the perpendicular from the centre bisects the chord,$AM = MB = AB / 2 = 16 / 2 = 8 \, cm$.
In right-angled triangle $\triangle PAM$,by Pythagoras theorem: $r^2 = PM^2 + AM^2 = 15^2 + 8^2 = 225 + 64 = 289$.
So,$r = \sqrt{289} = 17 \, cm$.
Now,in right-angled triangle $\triangle PNC$,by Pythagoras theorem: $r^2 = PN^2 + CN^2$.
$17^2 = 8^2 + CN^2 \implies 289 = 64 + CN^2$.
$CN^2 = 289 - 64 = 225$.
$CN = \sqrt{225} = 15 \, cm$.
Since $N$ is the midpoint of $CD$,$CD = 2 \times CN = 2 \times 15 = 30 \, cm$.

(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 $PM + PN = 24 + 10 = 34\,cm$.

(D) Let the centre of the circle be $P$ and the radius $r = 25 \, cm$.
Draw perpendiculars from $P$ to the chords $AB$ and $CD$,meeting them at points $M$ and $N$ respectively.
Since the perpendicular from the centre to a chord bisects the chord,we have:
$AM = MB = \frac{AB}{2} = \frac{48}{2} = 24 \, cm$
$CN = ND = \frac{CD}{2} = \frac{40}{2} = 20 \, cm$
In right-angled triangle $\triangle PMA$:
$PM^2 + AM^2 = PA^2$
$PM^2 + 24^2 = 25^2$
$PM^2 + 576 = 625$
$PM^2 = 49 \implies PM = 7 \, cm$
In right-angled triangle $\triangle PNC$:
$PN^2 + CN^2 = PC^2$
$PN^2 + 20^2 = 25^2$
$PN^2 + 400 = 625$
$PN^2 = 225 \implies PN = 15 \, cm$
Since the centre $P$ lies between the chords,the distance between $AB$ and $CD$ is $PM + PN = 7 \, cm + 15 \, cm = 22 \, cm$.

(A) Let $P$ be the center of the circle with radius $r = 13 \, cm$. Let $M$ and $N$ be the midpoints of chords $AB$ and $CD$ respectively. Since the perpendicular from the center to a chord bisects the chord,$AM = MB = 12 \, cm$ and $CN = ND$. In $\triangle PMA$,by the Pythagorean theorem,$PM^2 + AM^2 = PA^2$,so $PM^2 + 12^2 = 13^2$,which gives $PM^2 = 169 - 144 = 25$,so $PM = 5 \, cm$. There are two cases: chords are on the same side or opposite sides of the center. Case $1$: Chords are on opposite sides. The distance $MN = PM + PN = 17 \, cm$. Since $PM = 5 \, cm$,$PN = 17 - 5 = 12 \, cm$. In $\triangle PNC$,$PN^2 + CN^2 = PC^2$,so $12^2 + CN^2 = 13^2$,which gives $CN^2 = 169 - 144 = 25$,so $CN = 5 \, cm$. Thus,$CD = 2 \times CN = 10 \, cm$. Case $2$: Chords are on the same side. The distance $MN = |PM - PN| = 17 \, cm$. Since $PM = 5 \, cm$,$PN$ would be $22 \, cm$ or $-12 \, cm$,which is impossible as the radius is $13 \, cm$. Thus,$CD = 10 \, cm$.

(B) Let the radius $r = 50 \, cm$. Let $M$ and $N$ be the midpoints of chords $AB$ and $CD$ respectively.
Since $AB = 80 \, cm$,$AM = MB = 40 \, cm$.
In $\triangle PMA$,by Pythagoras theorem,$PM^2 + AM^2 = PA^2$.
$PM^2 + 40^2 = 50^2 \implies PM^2 = 2500 - 1600 = 900 \implies PM = 30 \, cm$.
Since $P$ does not lie between the chords,the distance between the chords is $PN - PM = 10 \, cm$.
$PN = 10 + PM = 10 + 30 = 40 \, cm$.
In $\triangle PNC$,$PN^2 + NC^2 = PC^2$.
$40^2 + NC^2 = 50^2 \implies 1600 + NC^2 = 2500 \implies NC^2 = 900 \implies NC = 30 \, cm$.
Since $N$ is the midpoint of $CD$,$CD = 2 \times NC = 2 \times 30 = 60 \, cm$.

(C) According to the circle theorem,the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
Here,the arc $AB$ subtends $\angle APB$ at the centre $P$ and $\angle ACB$ at the remaining part of the circle.
Therefore,$\angle APB = 2 \angle ACB$.
Given that $\angle ACB = 50^{\circ}$.
Substituting the value,we get $\angle APB = 2 \times 50^{\circ} = 100^{\circ}$.
Thus,the value of $\angle APB$ is $100^{\circ}$.

(D) Given that $AB$ is a chord of a circle with centre $P$. Point $C$ is a point on the major arc $AB$.
We are given that $\angle APB = 130^{\circ}$.
According to the circle theorem,the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
Therefore,$\angle APB = 2 \angle ACB$.
Substituting the given value:
$130^{\circ} = 2 \angle ACB$
$\angle ACB = \frac{130^{\circ}}{2} = 65^{\circ}$.
Thus,$\angle ACB = 65^{\circ}$.

(A) Given that $AB$ is a chord of a circle with centre $P$. Point $C$ is a point on the major arc $AB$.
According to the circle theorem,the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
Therefore,$\angle APB = 2 \angle ACB$.
Given that $\angle ACB + \angle APB = 135^{\circ}$.
Substituting $\angle APB = 2 \angle ACB$ in the given equation:
$\angle ACB + 2 \angle ACB = 135^{\circ}$
$3 \angle ACB = 135^{\circ}$
$\angle ACB = \frac{135^{\circ}}{3} = 45^{\circ}$.
Now,calculating $\angle APB$:
$\angle APB = 2 \times 45^{\circ} = 90^{\circ}$.
Thus,$\angle ACB = 45^{\circ}$ and $\angle APB = 90^{\circ}$.

(B) We know that angles in the same segment of a circle are equal.
Therefore,$\angle BDC = \angle BAC$.
Given $\angle BAC = 50^{\circ}$,so $\angle BDC = 50^{\circ}$.
Now,$\angle ADC = \angle ADB + \angle BDC$.
Substituting the values,$\angle ADC = 45^{\circ} + 50^{\circ} = 95^{\circ}$.
Since $ABCD$ is a cyclic quadrilateral,the sum of opposite angles is $180^{\circ}$.
Therefore,$\angle ABC + \angle ADC = 180^{\circ}$.
$\angle ABC + 95^{\circ} = 180^{\circ}$.
$\angle ABC = 180^{\circ} - 95^{\circ} = 85^{\circ}$.

(N/A) Let the sides $AB$ and $AC$ of $\Delta ABC$ be the diameters of two circles. These circles intersect each other at points $A$ and $P$.
Draw the common chord $AP$.
Since $AB$ is a diameter,$\angle APB$ is an angle in a semicircle.
$\therefore \angle APB = 90^{\circ}$.
Since $AC$ is a diameter,$\angle APC$ is an angle in a semicircle.
$\therefore \angle APC = 90^{\circ}$.
Adding these two equations,we get:
$\angle APB + \angle APC = 90^{\circ} + 90^{\circ} = 180^{\circ}$.
Since $\angle APB$ and $\angle APC$ are adjacent angles with a common arm $AP$ and their sum is $180^{\circ}$,they form a linear pair.
Therefore,the points $B, P,$ and $C$ must lie on a straight line.
Hence,the point of intersection $P$ of the circles lies on the third side $BC$ of the triangle.

(N/A) The bisectors of $\angle A, \angle B$ and $\angle C$ of $\Delta ABC$ intersect the circumcircle of $\Delta ABC$ at $D, E$ and $F$ respectively.
$\angle FDE = \angle FDA + \angle EDA$ (Adjacent angles)
$= \angle FCA + \angle EBA$ (Angles in the same segment of the circle are equal)
$= \frac{1}{2} \angle C + \frac{1}{2} \angle B$ (Since $AD, BE, CF$ are angle bisectors)
$= \frac{1}{2}(\angle B + \angle C)$
$= \frac{1}{2}(180^{\circ} - \angle A)$ (Since $\angle A + \angle B + \angle C = 180^{\circ}$)
$= 90^{\circ} - \frac{1}{2} \angle A$
Thus,$\angle FDE = 90^{\circ} - \frac{1}{2} \angle A$.
Similarly,$\angle DEF = 90^{\circ} - \frac{1}{2} \angle B$ and $\angle EFD = 90^{\circ} - \frac{1}{2} \angle C$.