Mix Examples - Circles Questions in English

(N/A) $(i)$ The centre of a circle always lies in the interior of the circle.
$(ii)$ If the distance of a point from the centre is greater than the radius,the point lies in the exterior of the circle.
$(iii)$ The longest chord of a circle is its diameter.
$(iv)$ An arc is a semicircle when its ends are the ends of a diameter.
$(v)$ $A$ segment of a circle is the region between an arc and the chord of the circle.
$(vi)$ $A$ circle divides the plane on which it lies into three parts: the interior,the circle itself,and the exterior.

(A-D) $(i)$ True. The definition of a radius is the line segment joining the centre to any point on the circle.
$(ii)$ False. $A$ circle can have an infinite number of equal chords of any given length.
$(iii)$ False. If a circle is divided into three equal arcs,each arc measures $120^{\circ}$,which is less than a semicircle $(180^{\circ})$,so each is a minor arc.
$(iv)$ True. By definition,the diameter is the longest chord of a circle and its length is equal to $2 \times \text{radius}$.
$(v)$ False. The region between a chord and its corresponding arc is called a segment,whereas a sector is the region between an arc and the two radii joining the centre to the endpoints of the arc.
$(vi)$ True. $A$ circle is a collection of all points in a plane that are at a fixed distance from a fixed point (the centre).

(A) Let the circle have its centre at $O'$. Let the chords $AC$ and $BD$ intersect at point $O$.
Since the chords bisect each other,$AO = OC$ and $BO = OD$.
In $\Delta AOB$ and $\Delta COD$:
$AO = OC$ (given),
$BO = OD$ (given),
$\angle AOB = \angle COD$ (vertically opposite angles).
Therefore,$\Delta AOB \cong \Delta COD$ by the $SAS$ congruence rule.
This implies $AB = CD$ and $\angle OAB = \angle OCD$. Since these are alternate interior angles,$AB \parallel CD$.
$A$ quadrilateral with one pair of opposite sides equal and parallel is a parallelogram. Thus,$ABCD$ is a parallelogram.
Since $ABCD$ is a cyclic quadrilateral (as its vertices lie on the circle),the sum of its opposite angles is $180^{\circ}$.
In a parallelogram,opposite angles are equal,so $\angle A = \angle C$ and $\angle B = \angle D$.
Since $\angle A + \angle C = 180^{\circ}$,we have $2\angle A = 180^{\circ}$,which means $\angle A = 90^{\circ}$.
$A$ parallelogram with one angle equal to $90^{\circ}$ is a rectangle. Therefore,$ABCD$ is a rectangle.

(A) The length of an arc of a circle is given by the formula $L = \frac{\theta}{360^{\circ}} \times 2\pi r$,where $\theta$ is the angle subtended by the arc at the center and $r$ is the radius of the circle.
Since the two circles are congruent,they have the same radius,let it be $r$.
Let $L_1$ be the length of arc $AXB$ and $L_2$ be the length of arc $A'YB'$.
Given that the angle subtended by arc $AXB$ at center $O$ is $\theta_1 = 75^{\circ}$.
So,$L_1 = \frac{75^{\circ}}{360^{\circ}} \times 2\pi r$.
Given that the angle subtended by arc $A'YB'$ at center $O'$ is $\theta_2 = 25^{\circ}$.
So,$L_2 = \frac{25^{\circ}}{360^{\circ}} \times 2\pi r$.
The ratio of the lengths of the arcs is $\frac{L_1}{L_2} = \frac{\frac{75^{\circ}}{360^{\circ}} \times 2\pi r}{\frac{25^{\circ}}{360^{\circ}} \times 2\pi r} = \frac{75^{\circ}}{25^{\circ}} = \frac{3}{1}$.
Therefore,the ratio is $3:1$.

(B) In $\triangle OPQ$,since $OP$ and $OQ$ are perpendiculars from the center to equal chords,$OP = OQ$.
Therefore,$\triangle OPQ$ is an isosceles triangle,which implies $\angle OQP = \angle OPQ = k$.
In $\triangle OPQ$,the sum of angles is $180^{\circ}$,so $\angle OQP + \angle OPQ + \angle POQ = 180^{\circ}$.
$k + k + 150^{\circ} = 180^{\circ}$.
$2k = 30^{\circ}$.
$k = 15^{\circ}$.
Thus,$\angle OPQ = 15^{\circ}$.
Since $OP \perp AB$,$\angle OPA = 90^{\circ}$.
From the figure,$\angle OPA = \angle OPQ + \angle APQ$.
$90^{\circ} = 15^{\circ} + \angle APQ$.
$\angle APQ = 90^{\circ} - 15^{\circ} = 75^{\circ}$.

(A) $AB$ is perpendicular to $BC$,therefore $\Delta ABC$ is a right-angled triangle with $\angle B = 90^{\circ}$.
Since the triangle is right-angled,the hypotenuse $AC$ is the diameter of the circle passing through points $A, B$ and $C$.
Using the Pythagorean theorem in $\Delta ABC$:
$AC = \sqrt{AB^2 + BC^2}$
$AC = \sqrt{12^2 + 16^2}$
$AC = \sqrt{144 + 256}$
$AC = \sqrt{400} = 20 \, cm$
Since the diameter $AC = 20 \, cm$,the radius $r = \frac{AC}{2} = \frac{20}{2} = 10 \, cm$.

(A) In $\Delta ADB$,the sum of angles is $180^{\circ}$.
$\angle DAB + \angle ABD + \angle ADB = 180^{\circ}$
$60^{\circ} + 50^{\circ} + \angle ADB = 180^{\circ}$
$110^{\circ} + \angle ADB = 180^{\circ}$
$\angle ADB = 180^{\circ} - 110^{\circ} = 70^{\circ}$
Since angles in the same segment of a circle are equal,we have $\angle ACB = \angle ADB$.
Therefore,$\angle ACB = 70^{\circ}$.

(D) In $\Delta OAB,$ we have $OA = OB$ (radii of the same circle).
Therefore,$\angle OAB = \angle OBA.$
Since the sum of angles in a triangle is $180^{\circ},$
$2 \angle OAB = 180^{\circ} - \angle AOB = 180^{\circ} - 90^{\circ} = 90^{\circ}.$
Thus,$\angle OAB = 45^{\circ}.$
Also,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.
So,$\angle ACB = \frac{1}{2} \angle AOB = \frac{1}{2} \times 90^{\circ} = 45^{\circ}.$
Now,in $\Delta CAB,$ the sum of angles is $180^{\circ}.$
$\angle CAB = 180^{\circ} - (\angle ABC + \angle ACB) = 180^{\circ} - (30^{\circ} + 45^{\circ}) = 180^{\circ} - 75^{\circ} = 105^{\circ}.$
Finally,$\angle CAO = \angle CAB - \angle OAB = 105^{\circ} - 45^{\circ} = 60^{\circ}.$

(B) The statement is False.
In $\triangle OAB$ and $\triangle OAC$,we have $OA = OA$ (common side) and $OB = OC$ (radii of the same circle).
However,the angles $\angle OAB$ and $\angle OAC$ are equal if and only if the chords $AB$ and $AC$ are equal in length $(AB = AC)$.
If the chords are of unequal lengths,the angles subtended by them at the center or the angles they make with the radius $OA$ will not be equal.
Therefore,the statement is not universally true.

(A) The given statement is True.
In two congruent circles,the radii are equal. Let the radius of both circles be $r$.
In $\triangle AOB$ and $\triangle AO^{\prime}B$:
$OA = O^{\prime}A = r$ (radii of congruent circles)
$OB = O^{\prime}B = r$ (radii of congruent circles)
$AB = AB$ (common chord)
By the $SSS$ (Side-Side-Side) congruence criterion,$\triangle AOB \cong \triangle AO^{\prime}B$.
Since the triangles are congruent,their corresponding angles are equal by $CPCT$ (Corresponding Parts of Congruent Triangles).
Therefore,$\angle AOB = \angle AO^{\prime}B$.

(B) The given statement is $False$. $A$ circle is uniquely determined by three non-collinear points. If three points are collinear,they lie on a single straight line. $A$ circle cannot pass through three points that lie on the same straight line because a circle has curvature,whereas a straight line does not. Therefore,it is impossible to draw a circle through three collinear points.

(A) The radius of the circle is $r = 3 \, cm$.
The diameter of the circle is $d = 2 \times r = 2 \times 3 \, cm = 6 \, cm$.
Since the distance between the two points $A$ and $B$ is $AB = 6 \, cm$,which is equal to the diameter of the circle,a circle can indeed be drawn passing through these two points with $AB$ as its diameter.
Therefore,the given statement is True.

(A) $AOB$ is a diameter of a circle and $C$ is a point on the circle.
Since the angle in a semicircle is a right angle,$\angle ACB = 90^{\circ}$.
In the right-angled triangle $\Delta ABC$,by applying the Pythagoras theorem:
$AC^{2} + BC^{2} = AB^{2}$.
Therefore,the given statement is True.

(B) The given statement is False.
Justification:
According to the theorem in geometry, 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.
If $D$ were the centre of the circle passing through $A, B,$ and $C$, then the angle subtended by the arc $BC$ at the centre $D$ $(\angle BDC)$ must be twice the angle subtended by the same arc at the circumference $(\angle BAC)$.
Given $\angle BAC = 30^{\circ}$, then $\angle BDC$ should be $2 \times 30^{\circ} = 60^{\circ}$.
However, the condition $\angle BDC = 60^{\circ}$ is necessary but not sufficient for $D$ to be the centre. There can be infinitely many points $D$ such that $\angle BDC = 60^{\circ}$ (for example, any point on the major arc $BC$ of a circle passing through $B, C$ and $D$).
Therefore, $D$ is not necessarily the centre of the circle passing through $A, B,$ and $C$.

(TRUE) The given statement is true.
According to the theorem,if a line segment joining two points subtends equal angles at two other points lying on the same side of the line containing the segment,then the four points lie on a circle (i.e.,they are concyclic).
Here,the line segment $BC$ subtends equal angles $\angle BAC = 45^{\circ}$ and $\angle BDC = 45^{\circ}$ at points $A$ and $D$ respectively.
Since these points lie on the same side of $BC$,the points $A, B, C$ and $D$ must be concyclic.

(TRUE) $AOB$ is a diameter of the circle with center $O$.
Consider the cyclic quadrilateral $ADCB$ (or $ADCEB$ as shown in the figure,but focusing on the cyclic quadrilateral $ADCB$ is sufficient).
Since $ADCB$ is a cyclic quadrilateral,the sum of opposite angles is $180^{\circ}$.
Therefore,$\angle ADC + \angle ABC = 180^{\circ}$.
Given $\angle ADC = 120^{\circ}$,we have $120^{\circ} + \angle ABC = 180^{\circ}$.
$\angle ABC = 180^{\circ} - 120^{\circ} = 60^{\circ}$.
In $\Delta ABC$,$\angle ACB = 90^{\circ}$ (Angle in a semicircle).
In $\Delta ABC$,the sum of angles is $180^{\circ}$.
$\angle CAB + \angle ABC + \angle ACB = 180^{\circ}$.
$\angle CAB + 60^{\circ} + 90^{\circ} = 180^{\circ}$.
$\angle CAB + 150^{\circ} = 180^{\circ}$.
$\angle CAB = 180^{\circ} - 150^{\circ} = 30^{\circ}$.
Hence,the given statement is True.

$(120^{\circ})$ We know that the angle subtended by an arc at the center is proportional to the length of the arc.
Given that $\operatorname{arc} AXB = \frac{1}{2} \operatorname{arc} BYC$,it follows that $\angle AOB = \frac{1}{2} \angle BOC$.
Since $AOC$ is a diameter,$\angle AOB + \angle BOC = 180^{\circ}$ (linear pair).
Substituting $\angle AOB = \frac{1}{2} \angle BOC$ into the equation,we get:
$\frac{1}{2} \angle BOC + \angle BOC = 180^{\circ}$
$\frac{3}{2} \angle BOC = 180^{\circ}$
$\angle BOC = 180^{\circ} \times \frac{2}{3} = 120^{\circ}$.

(A) We are given that arc $AXB \cong$ arc $CYD$.
According to the properties of circles,if two arcs of a circle are congruent,then their corresponding chords are equal.
Therefore,chord $AB =$ chord $CD$.
Thus,the ratio of the lengths of the chords is $AB:CD = 1:1$.

(N/A) Given: Three non-collinear points $A, B$ and $C$ lie on a circle.
To prove: The perpendicular bisectors of $AB, BC$ and $CA$ are concurrent.
Construction: Join $AB, BC$ and $CA$. Draw the perpendicular bisectors $ST$ of $AB$,$PM$ of $BC$,and $QR$ of $CA$. Since points $A, B$ and $C$ are not collinear,the lines $ST, PM$ and $QR$ are not parallel and will intersect at some point $O$.
Proof:
$1$. Since point $O$ lies on $ST$,which is the perpendicular bisector of $AB$,we have $OA = OB$ (Any point on the perpendicular bisector of a line segment is equidistant from its endpoints) ... $(1)$
$2$. Similarly,since $O$ lies on $PM$,the perpendicular bisector of $BC$,we have $OB = OC$ ... $(2)$
$3$. Also,since $O$ lies on $QR$,the perpendicular bisector of $CA$,we have $OC = OA$ ... $(3)$
From $(1)$,$(2)$,and $(3)$,we get $OA = OB = OC = r$ (where $r$ is the radius).
This implies that $O$ is the center of the circle passing through $A, B$ and $C$. Since the perpendicular bisectors of the sides of a triangle intersect at a unique point (the circumcenter),the perpendicular bisectors of $AB, BC$ and $CA$ must be concurrent at point $O$.

(N/A) To prove: The bisector $AM$ of $\angle BAC$ passes through the centre $O$.
Construction: Join $BC$. Let the bisector $AM$ intersect $BC$ at $P$.
Proof: In $\Delta BAP$ and $\Delta CAP$:
$AB = AC$ (Given,as chords are equal)
$\angle BAP = \angle CAP$ (Given,as $AM$ is the bisector)
$AP = AP$ (Common side)
Therefore,$\Delta BAP \cong \Delta CAP$ (By $SAS$ congruence rule).
Thus,$BP = CP$ and $\angle BPA = \angle CPA$ (by $CPCT$).
Since $\angle BPA + \angle CPA = 180^{\circ}$ (Linear pair),
$\angle BPA = \angle CPA = 90^{\circ}$.
This means $AP$ is the perpendicular bisector of the chord $BC$. We know that the perpendicular bisector of any chord of a circle always passes through the centre of the circle. Hence,the bisector of $\angle BAC$ passes through the centre $O$.

(N/A) $ABCD$ is a quadrilateral such that $A$ is the centre of the circle passing through $B, C$ and $D$. We have to prove that $\angle CBD + \angle CDB = \frac{1}{2} \angle BAD$.
Join $AC$.
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:
Therefore,$\angle CAD = 2 \angle CBD$ $....(1)$
And $\angle BAC = 2 \angle CDB$ $....(2)$
Adding $(1)$ and $(2)$,we get:
$\angle CAD + \angle BAC = 2(\angle CBD + \angle CDB)$
$\Rightarrow \angle BAD = 2(\angle CBD + \angle CDB)$
Hence,$\angle CBD + \angle CDB = \frac{1}{2} \angle BAD$.

(N/A) Given: $O$ is the circumcentre of $\Delta ABC$ and $D$ is the mid-point of $BC$,so $OD \perp BC$.
To prove: $\angle BOD = \angle A$.
Construction: Join $OB$ and $OC$.
Proof: In $\Delta OBD$ and $\Delta OCD$,we have:
$OB = OC$ (Radii of the same circumcircle)
$OD = OD$ (Common side)
$BD = CD$ ($D$ is the mid-point of $BC$)
Therefore,$\Delta OBD \cong \Delta OCD$ by $SSS$ congruence rule.
This implies $\angle BOD = \angle COD$ (by $CPCT$).
Thus,$\angle BOC = \angle BOD + \angle COD = 2\angle BOD$.
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.
Therefore,$\angle BOC = 2\angle BAC = 2\angle A$.
Equating the two expressions for $\angle BOC$:
$2\angle BOD = 2\angle A$
$\angle BOD = \angle A$.
Hence,proved.

$(120^{\circ})$ Given that chords $AB$ and $AC$ subtend angles $\angle AOB = 150^{\circ}$ and $\angle AOC = 90^{\circ}$ at the centre $O$.
Since $AB$ and $AC$ lie on opposite sides of the centre,the angle subtended by the chord $BC$ at the centre is $\angle BOC = \angle AOB + \angle AOC = 150^{\circ} + 90^{\circ} = 240^{\circ}$.
The angle subtended by the chord $BC$ at the remaining part of the circle is $\angle BAC$.
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.
Therefore,the reflex angle $\angle BOC = 2 \angle BAC$.
$240^{\circ} = 2 \angle BAC$.
$\angle BAC = \frac{240^{\circ}}{2} = 120^{\circ}$.

(N/A) $ABC$ is a triangle and $O$ is the circumcentre.
Draw $OD \perp BC$. Join $OB$ and $OC$.
In right $\Delta OBD$ and right $\Delta OCD$,we have:
hyp. $OB = \text{hyp. } OC$ [Radii of the same circle]
$OD = OD$ [Common side]
$\therefore \Delta OBD \cong \Delta OCD$ [By $RHS$ congruence rule]
$\therefore \angle 1 = \angle 2$ and $\angle 3 = \angle 4$ [$CPCT$]
Now,$\angle BOC = 2 \angle BAC$ [Angle at the centre is double the angle at the circumference]
Also,in $\Delta OBC$,$OB = OC$,so $\angle 3 = \angle 4$.
In $\Delta OBD$,$\angle 3 + \angle 1 + 90^{\circ} = 180^{\circ} \Rightarrow \angle 3 + \angle 1 = 90^{\circ}$.
Since $\angle BOC = 2 \angle 1 + 2 \angle 2$ and $\angle BOC = 2 \angle BAC$,we have $\angle 1 + \angle 2 = \angle BAC$.
Since $\angle 1 = \angle 2$,we have $2 \angle 1 = \angle BAC$,or $\angle 1 = \frac{1}{2} \angle BAC$.
Substituting this into $\angle 3 + \angle 1 = 90^{\circ}$,we get $\angle OBC + \angle BAC = 90^{\circ}$ (where $\angle OBC = \angle 3$).
Hence,proved.

$(270^{\circ})$ Join $BC$.
Since the angle in a semicircle is $90^{\circ}$,we have
$\angle ACB = 90^{\circ}$.
As $ACDEB$ is a cyclic quadrilateral (all vertices lie on the circle),the sum of opposite angles is $180^{\circ}$.
Therefore,$\angle BCD + \angle BED = 180^{\circ}$.
Now,adding $\angle ACB$ to both sides,we get:
$(\angle BCD + \angle ACB) + \angle BED = 180^{\circ} + \angle ACB$
Since $\angle BCD + \angle ACB = \angle ACD$,we have:
$\angle ACD + \angle BED = 180^{\circ} + 90^{\circ} = 270^{\circ}$.

(N/A) In $\Delta OBC$,we have:
$OB = OC$ (Radii of the same circle)
Therefore,$\angle OCB = \angle OBC = 57^{\circ}$ (Since $\angle OCB = 57^{\circ}$ is given,and angles opposite to equal sides are equal).
Now,in $\Delta BOC$,the sum of angles is $180^{\circ}$:
$\angle OCB + \angle OBC + \angle BOC = 180^{\circ}$
$57^{\circ} + 57^{\circ} + \angle BOC = 180^{\circ}$
$114^{\circ} + \angle BOC = 180^{\circ}$
$\angle BOC = 180^{\circ} - 114^{\circ} = 66^{\circ}$.
In $\Delta OAB$,we have:
$OA = OB$ (Radii of the same circle)
Therefore,$\angle OBA = \angle OAB = 30^{\circ}$.
In $\Delta OAB$,the sum of angles is $180^{\circ}$:
$\angle OAB + \angle OBA + \angle AOB = 180^{\circ}$
$30^{\circ} + 30^{\circ} + \angle AOB = 180^{\circ}$
$\angle AOB = 180^{\circ} - 60^{\circ} = 120^{\circ}$.
Since $\angle AOB = \angle AOC + \angle BOC$,we have:
$120^{\circ} = \angle AOC + 66^{\circ}$
$\angle AOC = 120^{\circ} - 66^{\circ} = 54^{\circ}$.
Hence,$\angle BOC = 66^{\circ}$ and $\angle AOC = 54^{\circ}$.

(N/A) Assume that two circles intersect at three distinct points,$A$,$B$,and $C$.
Since $A$,$B$,and $C$ are three distinct points,they are non-collinear.
According to the geometric theorem,there is one and only one circle that passes through three non-collinear points.
Therefore,if two circles pass through the same three points $A$,$B$,and $C$,they must be the same circle.
This contradicts the assumption that there are two distinct circles.
Hence,it is impossible for two distinct circles to intersect at more than two points.

(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 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.

(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 $RS = 6\, cm$.

(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).

(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,$AB = AC$.

(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 $SD = 2x = 2(10\sqrt{3}) = 20\sqrt{3} \, m$.

(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 $\angle EFD = 90^{\circ} - \frac{1}{2} \angle C$.

(N/A) $(1)$ False. $A$ line segment joining any two points of a circle is called a chord. $A$ diameter is a special type of chord that passes through the center of the circle.
$(2)$ True. By definition, the diameter is the longest chord of a circle and its length is exactly twice the radius $(d = 2r)$.
$(3)$ False. The longest chord of a circle is the diameter. For a circle with radius $r = 14 \text{ cm}$, the diameter is $d = 2 \times 14 \text{ cm} = 28 \text{ cm}$. Since the length of any chord cannot exceed the diameter, a chord of length $32 \text{ cm}$ is impossible in this circle.

(A) $(1)$ False. There is only one unique circle that passes through three given non-collinear points.
$(2)$ True. Infinitely many circles can be drawn passing through two given points,as their centers can lie anywhere on the perpendicular bisector of the line segment joining the two points.

(B) When two circles intersect at two distinct points,the line segment joining these two points is called a common chord. Since two circles can intersect at a maximum of two points,they can have only one common chord connecting those two points. Therefore,the maximum number of common chords for two circles in the same plane is $1$.