**NCERT Solutions Class 10 Maths Chapter 10 Circles** – Here are all the NCERT solutions for Class 10 Maths Chapter 10. This solution contains questions, answers, images, explanations of the complete Chapter 10 titled Circles of Maths taught in Class 10. If you are a student of Class 10 who is using NCERT Textbook to study Maths, then you must come across Chapter 10 Circles. After you have studied lesson, you must be looking for answers of its questions. Here you can get complete NCERT Solutions for Class 10 Maths Chapter 10 Circles in one place.

## NCERT Solutions Class 9 Maths Chapter 10 Circles

Here on **AglaSem Schools**, you can access to **NCERT Book Solutions** in free pdf for Maths for Class 9 so that you can refer them as and when required. The NCERT Solutions to the questions after every unit of NCERT textbooks aimed at helping students solving difficult questions.

For a better understanding of this chapter, you should also see summary of Chapter 10 Circles , Maths, Class 9.

Class | 9 |

Subject | Maths |

Book | Mathematics |

Chapter Number | 10 |

Chapter Name |
Circles |

### NCERT Solutions Class 9 Maths chapter 10 Circles

Class 9, Maths chapter 10, Circles solutions are given below in PDF format. You can view them online or download PDF file for future use.

### Circles

Q.1:Fill in the blanks: (i) The centre of a circle lies in_ of the circle. (exterior/ interior) (ii) A point, whose distance from the centre of a circle is greater than its radius lies in _of the circle. (exterior/ interior) (iii) The longest chord of a circle is a _ of the circle. (iv) An arc is a _ when its ends are the ends of a diameter. (v) Segment of a circle is the region between an arc and _ of the circle. (vi) A circle divides the plane, on which it lies, in _ parts.

Ans :(i) The centre of a circle lies ininteriorof the circle. (ii) A point, whose distance from the centre Of a circle is greater than its radius lies Inexteriorof the circle. (iii) The longest chord of a circle is adiameterof the circle. (iv) An arc is a semi-circle when its ends are the ends of a diameter. (v) Segment of a circle is the region between an arc andchordof the circle. (vi) A circle divides the plane, on which it lies, inthreeparts.

Q.2:Write True or False: Give reasons for your answers. (i) Line segment joining the centre to any point on the circle is a radius of the circle. (ii) A circle has only finite number of equal chords. (iii) If a circle is divided into three equal arcs, each is a major arc. (iv) A chord of a circle, which is twice as long as its radius, is a diameter of the circle. (v) Sector is the region between the chord and its corresponding arc. (vi) A circle is a plane figure

Ans :(i) True. All the points on the circle are at equal distances from the centre of the circle, and this equal distance is called as radius of the circle. (ii) False. There are infinite points on a circle. Therefore, we can draw infinite number of chords of given length. Hence, a circle has infinite number of equalchords. (iii) false. Consider three arcs of same length as AB, BC, and CA. It can be observed that for minor arc BOC, CAB is a major arc. Therefore, A3, BC, and CA are minor arcs of the circle. (iv) True. Let AB be a chord which is twice as long as its radius. It can be observed that in this situation, our chord will be passing through the centre of the circle. Therefore, it will be the diameter of the circle. (v) False. Sector is the region between an arc and two radii joining the centre to the end points of the arc. For example, in the given figure, OAB is the sector of the circle. (vi) True. A circle is a two-dimensional figure and it can also be referred to as a plane figure.

Q.1:Recall that two circles are congruent if they have the same radii. Prove that equal chords of congruent circles subtend equal angles at their centres.

Ans :A circle is a collection of points which are equidistant from a fixed point. This fixed point is called as the centre of the circle and this equal distance is called as radius of the circle. And thus, the shape of a circle depends on its radius. Therefore, it can be observed that if we try to superimpose two circles of equal radius, then both circles will cover each other. Therefore, two circles are congruent if they have equal radius. Consider two congruent circles having centre O and O' and two chords AB and CD of equal lengths. In \( \triangle \mathrm{AOB}\) and \( \triangle \mathrm{CO'D}\) AB = CD (Chords of same length) OA = O'C (Radii of congruent circles) 0B = O'D (Radii of congruent circles) \( \triangle \mathrm{AOB}\) \( \triangle \mathrm{CO'D}\) (SSS congruence rule) \( \angle A O B=\angle C O^{\prime} D\) (By CPCT) Hence, equal chords of congruent circles subtend equal angles at their centres.

Q.2:Prove that if chords of congruent circles subtend equal angles at their centres, then the chords are equal.

Ans :In \( \triangle \mathrm{AOB}\) and \( \triangle \mathrm{CO'D}\) \( \angle A O B=\angle C O^{\prime} D\) (given) OB = O'D (Chords of same length) OA = O'C (Radii of congruent circles) \( \triangle \mathrm{AOB}\) \( \triangle \mathrm{CO'D}\) (SSS congruence rule) AB=CD (By CPCT) Hence , if the chord of congruent circle equal angles at their centers, then the chords are equals.

Q.1:Draw different pairs of circles. How many points does each pair have in common? What is the maximum number of common points?

Ans :Consider of the following pair of circles. The above circles do not intersect each other at any point . therefore, they do not have any point in common. The above circles touch each other only at one point Y. therefore, there is 1 point in common. The above circles touch each other at 1 point X only. Therefore the circles have one point in common. These circles intersect each other at two points G and H. Therefore, the circles have two points in common. It can be observed that there can be a maximum of 2 points in common. Consider the situation in which two congruent circles are superimposed on each other. This situation can be referred to as if we are drawing the circle two times.

Q.2:Suppose you are given a circle. Give a construction to find its centre.

Ans :The below given steps will be followed to find the centre of the given circle. Step1. Take the given circle. Step2. Take any two different chords Ad and CD of this circle and draw perpendicular bisectors of these chords. Step3. Let these perpendicular bisectors meet at point O. Hence, O is the centre of the given circle.

Q.3:If two circles intersect at two points, prove that their centres lie on the perpendicular bisector of the common chord.

Ans :Consider two circles centered at point O and O', intersecting each other at point A and B respectively. Join AB. AB is the chord of the circle centered at O. Therefore, perpendicular bisector of AB will pass through O. Again, AB is also the chord of the circle centered at O'. Therefore, perpendicular bisector of AB will also pass through O'. Clearly, the centres of these circles lie on the perpendicular bisector of the common chord.

Q.1:Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centres is 4 cm. Find the length of the common chord.

Ans :Let the radius of the circle centered at O and O' be 5 cm and 3 cm respectively. OA = OB=5 cm O'A= O'B = 3 cm 00' will be the perpendicular bisector of chord AB. AC=CB It is given that, 00' = 4 cm Let OC be x. Therefore, O'C will be 4 - x . In\( \triangle \mathrm{O} \mathrm{A} \mathrm{C}_{1}\) \( \mathrm{OA}^{2}=\mathrm{AC}^{2}+\mathrm{OC}^{2}\) \( \mathrm{5}^{2}=\mathrm{AC}^{2}+\mathrm{x}^{2}\) \( \Rightarrow 25-x^{2}=\mathrm{AC}^{2} \).........(1) In \( \triangle \mathrm{O}^{\prime} \mathrm{A} \mathrm{C}_{1}\) \( \mathrm{O'A}^{2}=\mathrm{AC}^{2}+\mathrm{O'C}^{2}\) \( \mathrm{3}^{2}=\mathrm{AC}^{2}+\mathrm{(4-x)}^{2}\) \( 9=A C^{2}+16+x^{2}-8 x\) \( A C^{2}=-x^{2}-7+8 x\).......(2) From equations (1) and (2) , we obtain \(\ 25-x^{2}=-x^{2}-7+8 x\) 8X=32 x=4 Therefore, the common chord will pass through the centre of the smaller circle i.e., O' and hence, it will be the diameter of the smaller circle. \( A C^{2}=25-x^{2}=25-4^{2}=25-16=9\) AC= 3m Length of the common chord AB = 2AC =( 2x3 )m = 6m

Q.2:If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to corresponding segments of the other chord.

Ans :Let PQ and RS be two equal chords of a given circle and they are intersecting each other at point T. Draw perpendiculars OV and OU on these chords. In \( \Delta O V T\) and \( Delta O U T\) OV =OU (Equal chords of a circle are equidistant from the centre) \( \angle \mathrm{OVT}=\angle \mathrm{OUT}\) (each 90 degree) OT=OT (common) \( \Delta O V T \equiv \Delta O U T \) ( RHS congruence rule) VT=UT (By CPCT)......(1) It is given that, PQ=RS……..(2) \( \frac{1}{2} \mathrm{PQ}=\frac{1}{2} \mathrm{RS}\) PV=RU……..(3) On adding equation (1) and (3) , we obtain PV + VT = RU + UT PT= RT……(4) On subtracting equation (4) from equation (2),we obtain PQ - PT = RU + UT QT = ST……(5) Equations (4) and (5) indicate that the corresponding segments Of chords PQ and RS are congruent to each other.

Q.3:If two equal chords of a circle intersect within the circle, prove that the line joining the point of intersection to the centre makes equal angles with the chords.

Ans :Let PQ and RS are two equal chords of a given circle and they are intersecting each other at point T. Draw perpendiculars OV and OU on these chords. In \( \Delta O V T\) and \( \Delta O U T\) OV = OU(Equal chords of a circle are equidistant from the centre) \( \angle \mathrm{OVT}=\angle \mathrm{OUT}\) (each 90 degree) OT=OT (common) \( \Delta O V T=\Delta O U T\) ( RHS congruence rule) \( \Delta O T V=\Delta O T U\) Therefore, it is proved that the line joining the point Of intersection to the centre makes equal angles with the chords.

Q.4:If a line intersects two concentric circles (circles with the same centre) with centre O at A, B, C and D, prove that AB = CD (see Fig. 10.25).

Ans :Let us draw a perpendicular OM on line AD. It can be observed that BC is the chord Of the smaller circle and AD is the chord Of the bigger circle. We know that perpendicular drawn from the centre of the circle bisects the chord. \( \begin{array}{l}{ \mathrm{BM}=\mathrm{MC} \ldots(1)} \\ {\mathrm{And}, \mathrm{AM}=\mathrm{MD} \ldots(2)}\end{array}\) On subtracting equation (2) from (1), we obtain AM - BM = MD - MC \( A B=C D\)

Q.5:Three girls Reshma, Salma and Mandip are playing a game by standing on a circle of radius 5m drawn in a park. Reshma throws a ball to Salma, Salma to Mandip, Mandip to Reshma. If the distance between Reshma and Salma and between Salma and Mandip is 6m each, what is the distance between Reshma and Mandip?

Ans :Draw perpendiculars OA and 03 on RS and SM respectively. \( \mathrm{AR}=\mathrm{AS}=\frac{6}{2}=3 \mathrm{m}\) OR =OS OM =5m (radi of the circle) In \( \Delta O A R_{1}\) \( \mathrm{OA}^{2}+\mathrm{AR}^{2}=\mathrm{OR}^{2}\) \( \mathrm{OA}^{2}+(3 \mathrm{m})^{2}=(5 \mathrm{m})^{2}\) \( \mathrm{OA}^{2}=(25-9) \mathrm{m}^{2}=16 \mathrm{m}^{2}\) OA =4m { ORSM will be a kite } \( \mathrm{OR}=\mathrm{OM} \text { and } \mathrm{RS}=\mathrm{SM} \). We know that the diagonals of a kite are perpendicular and the diagonal common to both the isosceles triangles is bisected by another diagonal. \( \Delta R C S \text { will be of } 90^{\circ} \text { and } R C=C M\) Area of \( \Delta \mathrm{ORS}=\frac{1}{2} \times \mathrm{OA} \times \mathrm{RS}\) \( \frac{1}{2} \times \mathrm{RC} \times \mathrm{OS}=\frac{1}{2} \times 4 \times 6\) RC x 5 = 24 RC = 4.8 RM = 2 RC = 2(4.8) = 9.6 Therefore, the distance between Reshma and Mandip is 9.6 m.

Q.6:A circular park of radius 20m is situated in a colony. Three boys Ankur, Syed and David are sitting at equal distance on its boundary each having a toy telephone in his hands to talk each other. Find the length of the string of each phone.

Ans :It iS Given that AS = SD - DA Therefore, \( \triangle \mathrm{ASD}\) is an equilateral triangle. OA (radius) = 20 m Medians of equilateral triangle pass through the circum entre (O) of the equilateral triangle ASD. We also know that medians intersect each other in the ratio 2: 1. As AB is the median of equilateral triangle ASD, we can write \( \Rightarrow \frac{\mathrm{OA}}{\mathrm{OB}}=\frac{2}{1}\) \( \Rightarrow \frac{20 \mathrm{m}}{\mathrm{OB}}=\frac{2}{1}\) \( \Rightarrow \mathrm{OB}=\left(\frac{20}{2}\right) \mathrm{m}=10 \mathrm{m}\) \( \angle AB=OA+OB=(20+10) m=30 \mathrm{m}\) \( \triangle \mathrm{ABD}\) , \( \mathrm{AD}^{2}=\mathrm{AB}^{2}+\mathrm{BD}^{2}\) \( A D^{2}=(30)^{2}+\left(\frac{A D}{2}\right)^{2}\) \( \mathrm{AD}^{2}=900+\frac{1}{4} \mathrm{AD}^{2}\) \( \frac{3}{4} \mathrm{AD}^{2}=900\) \( \mathrm{AD}^{2}=1200\) \( \mathrm{AD}=20 \sqrt{3}\) Therefore , the length of the string of each phone will be \( 20 \sqrt{3} \mathrm{m}\)

Q.1:In Fig, A,B and C are three points on a circle with centre O such that ∠ BOC = 30° and ∠ AOB = 60°. If D is a point on the circle other than the arc ABC, find ∠ADC.

Ans :It can observed that \( \angle A O C=\angle A O B + \angle B O C\) \(b \begin{array}{l}{=60^{\circ}+30^{\circ}} \\ {=90^{\circ}}\end{array}\) We know that angle subtended by an arc at the centre is double the angle subtended by it any point on the remaining part of the circle. \( \angle \mathrm{ADC}=\frac{1}{2} \angle \mathrm{AOC}=\frac{1}{2} \times 90^{\circ}=45^{\circ}\)

Q.2:A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.

Ans :In \( \triangle O A B\) OA =OB =AB (radius) \( \triangle \mathrm{O} \mathrm{A} \mathrm{B}\) is an equilateral triangles. \( \angle A O B=60^{\circ}\) \( \angle \mathrm{ACB}=\frac{1}{2} \angle \mathrm{AOB}=\frac{1}{2}\left(60^{\circ}\right)=30^{\circ}\) In cyclic equilateral ABCD, \( \angle A C B+\angle A D B=180^{\circ}\) (opposite angle in cyclic quadrilateral) \( \angle A D B=180^{\circ}-30^{\circ}=150^{\circ}\) Therefore , angle subtended by this chord at a point on the major arc and the minor arc are \( 30^{\circ} \text { and } 150^{\circ} \) respectively

Q.3:In Fig, ∠ PQR = 100°, where P, Q and R are points on a circle with centre O. Find ∠ OPR.

Ans :Consider PR as a chord of the circle. Take any point S on the major arc of the circle. PQRS is a cyclic quadrilateral. \( \begin{array}{l}{\angle \mathrm{PQR}+\angle \mathrm{PSR}=180^{\circ}} \\ {\angle \mathrm{PSR}=180^{\circ}-100^{\circ}}\end{array}\)=80 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. \( \angle P O R=2 \angle P S R=2\left(80^{\circ}\right)=160^{\circ}\) In \( \triangle \mathrm{POR}\) OP=OR (radii of the same circle ) \( \angle O P R=\angle O R P\) (Angles opposite to equal sides of a triangle) \( 2 \angle O P R+160^{\circ}=180^{\circ}\) \( 2 \angle O P R=180^{\circ}-160^{\circ}=20^{\circ}\) \( \angle \mathrm{OPR}=10^{\circ}\)

Q.4:In Fig, ∠ ABC = 69°, ∠ ACB = 31°, find ∠ BDC.

Ans :In \( \Delta \mathrm{ABC}, \angle \mathrm{ABC}=69^{\circ}, \angle \mathrm{ACB}=31^{\circ}\) So, \( \angle B A C=180^{\circ}-\left(69^{\circ}+31^{\circ}\right)\) \( =180^{\circ}-100^{\circ}\) \( =80^{\circ}\) \( \angle B D C=\angle B A C=80^{\circ}\) (at same are BC)

Q.5:In Fig, A, B, C and D are four points on a circle. AC and BD intersect at a point E such that ∠ BEC = 130° and ∠ ECD = 20°. Find ∠ BAC.

Ans :In \( \triangle \mathrm{CDE}\) \( \angle \mathrm{CDE}+\angle \mathrm{DCE}=\angle \mathrm{CEB}\)(exterior angle \( \angle C D E+20^{\circ}=130^{\circ}\) \( \angle C D E=110^{\circ}\) However , \( \angle B A C=\angle C D E\) (angles in the same segment of a circle) \( \angle B A C=110^{\circ}\)

Q.6:ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If ∠ DBC = 70°, ∠ BAC is 30°, find ∠ BCD. Further, if AB = BC, find ∠ ECD.

Ans :\( \angle \mathrm{CBD}=\angle \mathrm{CAD}\) \( \angle C A D=70^{\circ}\) \( \angle B A D=D B A C+\angle C A D=30^{\circ}+70^{\circ}=100^{\circ}\) \( \angle B C D+D B A D=180^{\circ}\) \( \angle B C D+100^{\circ}=180^{\circ}\) \( \angle B C D=80^{\circ}\) In \( \triangle \mathrm{ABC}\) \( A B=B C(\text { Given })\) \( \angle B C A=\angle C A B\) \( \angle B C A=30^{\circ}\) We have \( \angle B C D=80^{\circ}\) \( \angle B C A+\angle A C D=80^{\circ}\) \( 30^{\circ}+\angle A C D=80^{\circ}\) \( \angle A C D=50^{\circ}\) \( \angle E C D=50^{\circ}\)

Q.7:If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.

Ans :Let ABCD be a cyclic quadrilateral having diagonals BD and AC, intersecting each other at point O. \( \angle B A D=\frac{1}{2} \angle B O D=\frac{180^{\circ}}{2}=90^{\circ}\)(consider AC is chord) \( \angle A D C+\angle A B C=180^{\circ}\) (cyclic quadrilateral) \( 90^{\circ}+\angle A B C=180^{\circ}\) \( \angle A B C=90^{\circ}\) Each interior angle of a cyclic quadrilateral is of 90 degree. Hence, it is a rectangle.

Q.8:If the non-parallel sides of a trapezium are equal, prove that it is cyclic.

Ans :Consider a trapezium ABCD with AB I ICD and BC AD. \( \angle \mathrm{CD} \text { and } \mathrm{BN} \angle \mathrm{CD} .\) AM =BM (perpendicular distance between two parallel lines is same) \( \triangle A M D \triangle B N C\) (RHS congruence rules) \( \angle A D C=\angle B C D\) (CPCT)..........(1) \( \angle B A D \text { and } \angle A D C\) are on the same side of transversal AD \( \angle B A D+\angle A D C=180^{\circ}\)........(2) \( \angle B A D+\angle B C D=180^{\circ}\) [using equation …..(1)] This equation shows that the opposite angles are supplementary. Therefore, ABCD is a cyclic quadrilateral.

Q.9:Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D and P, Q respectively (see Fig. 10.40). Prove that ∠ACP = ∠ QCD.

Ans :Join chords AP and DQ. For chord AP, \( \angle P B A=\angle A C P\) (Angles in the same segment) …….(1) For chord DQ, \( \angle D B Q=\angle Q C D\) (Angles in the same segment) …….(2) ABD and PBQ are line segments intersecting at B. \( \angle P B A=\angle A C P\) (Angles in the same segment) …….(3) From equation (1) ,(2) ,and (3),we obtain \( \angle A C P=\angle Q C D\)

Q.10:If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lie on the third side.

Ans :consider a \( \triangle \mathrm{ABC}\). Two circles are drawn while taking AB and AC as the diameter. Let they intersect each other at D and let D not lie on BC. join AD. \( \angle A D B=90^{\circ}\) (angle subtended by semi - circle) \( \angle A D C=90^{\circ}\) ( angle subtended by semi - circle) \( \angle B D C=\angle A D B+\angle A D C=90^{\circ}+90^{\circ}=180^{\circ}\) \( \begin{array}{l}{\text { Therefore, } B D C \text { is a straight line and hence, our assumption was wrong. }} \\ {\text { Thus, Point D lies on third side BC of } \triangle A B C \text { . }}\end{array}\)

Q.11:ABC and ADC are two right triangles with common hypotenuse AC. Prove that ∠ CAD = ∠ CBD.

Ans :\( \triangle \mathrm{ABC}\) \( \angle A B C+\angle B C A+\angle C A B=180^{\circ}\) ( Angle sum property of a triangle) \( \angle 90^{\circ}+\angle B C A+\angle C A B=180^{\circ}\) \( \angle B C A+\angle C A B=90^{\circ}\)........(1) \( \triangle \mathrm{ADC}\) \( \angle C D A+\angle A C D+\angle D A C=180^{\circ}\)( Angle sum property of a trinangle) \( \angle 90^{\circ}+\angle A C D+\angle D A C=180^{\circ}\) \( \angle A C D+\angle D A C=90^{\circ}\)........(2) Adding equation (1) and (2) ,we obtain \( \angle B C A+\angle C A B+\angle A C D+\angle D A C=180^{\circ}\) \( \angle(\angle B C A+\angle A C D)+(\angle C A B+\angle D A C)=180^{\circ}\) \( \angle B C D+\angle D A B=180^{\circ}\).......(3) However , it is given that \( \angle B+\angle D=90^{\circ}+90^{\circ}=180^{\circ}\).......(4) From equations (3) and (4), it can be observed that the sum of the measures of opposite angles Of quadrilateral A3CD is 1800. Therefore, it is a cyclic quadrilateral. Consider chord CD. \( \angle C A D=\angle C B D\) (angle in the same segment )

Q.12:Prove that a cyclic parallelogram is a rectangle.

Ans :Let ABCD be a cyclic parallelogram. \( \angle A+\angle C=180^{\circ}\) We know that opposite angles Of a parallelogram are equal. \( \angle A=\angle C \text { and } \angle B=\angle D\) From equation (1), \( \angle A+\angle C=180^{\circ}\) \( \angle A+\angle A=180^{\circ}\) \( 2 \angle A=180^{\circ}\) \( \angle A=90^{\circ}\) Parallelogram ABCD has one of its interior angles as 900. Therefore, it is a rectangle.

Q.1:Prove that the line of centres of two intersecting circles subtends equal angles at the two points of intersection.

Ans :Let two circles having their centres as O and O' intersect each other at point A and B respectively. Let us join O O' In \( \Delta A O O^{\prime} \text { and } B O O^{\prime}\) OA =OB O'A = O'B OO' = OO' (common) \( \triangle \mathrm{AO}{\mathrm{O}^{\prime}} \triangle\mathrm{BO}{\mathrm{O}^{\prime}}\) (by CPCT) Therefore, line of centres of two intersecting circles subtends equal angles at the two points of intersection.

Q.2:Two chords AB and CD of lengths 5 cm and 11 cm respectively of a circle are parallel to each other and are on opposite sides of its centre. If the distance between AB and CD is 6 cm, find the radius of the circle.

Ans :Draw OM \(\angle AB\) and ON \(\angle CD\) . join OB and OD \( \mathrm{BM}=\frac{\mathrm{AB}}{2}=\frac{5}{2}\) \( \mathrm{ND}=\frac{\mathrm{CD}}{2}=\frac{11}{2}\) Let ON be x. Therefore, 0M will be 6 - x. In \( \triangle \mathrm{MOB}\) \( \mathrm{OM}^{2}+\mathrm{MB}^{2}=\mathrm{OB}^{2}\) \( (6-x)^{2}+\left(\frac{5}{2}\right)^{2}=\mathrm{OB}^{2}\) \( 36+x^{2}-12 x+\frac{25}{4}=\mathrm{OB}^{2}\)................(1) In \( \triangle \mathrm{NOD}\) \( \mathrm{ON}^{2}+\mathrm{ND}^{2}=\mathrm{OD}^{2}\) \( x^{2}+\left(\frac{11}{2}\right)^{2}=\mathrm{OD}^{2}\) \( x^{2}+\frac{121}{4}=\mathrm{OD}^{2}\)..............(2) We have 0B = OD (Radii of the same circle) Therefore, from equation (1) and (2), \( 36+x^{2}-12 x+\frac{25}{4}=x^{2}+\frac{121}{4}\) \( 12 x=36+\frac{25}{4}-\frac{121}{4}\) \( =\frac{144+25-121}{4}=\frac{48}{4}=12\) x=1 From equation (2), \( (1)^{2}+\left(\frac{121}{4}\right)=\mathrm{OD}^{2}\) \( \mathrm{OD}^{2}=1+\frac{121}{4}=\frac{125}{4}\) \( \mathrm{OD}=\frac{5}{2} \sqrt{5}\) Therefore , the radius of circle is \(\frac{5}{2} \sqrt{5}\) cm.

Q.3:The lengths of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chord is at distance 4 cm from the centre, what is the distance of the other chord from the centre?

Ans :Let AB and CD be two parallel chords in a circle centered at O. Join 03 and 00. Distance of smaller chord AB from the centre of the circle = 4 cm ,0M = 4 cm \( \mathrm{MB}=\frac{\mathrm{AB}}{2}=\frac{6}{2}=3 \mathrm{cm}\) In \( \triangle O M B\) \( \mathrm{OM}^{2}+\mathrm{MB}^{2}=\mathrm{OB}^{2}\) \( (4)^{2}+(3)^{2}=\mathrm{OB}^{2}\) \( 16+9=\mathrm{OB}^{2}\) \( \mathrm{OB}=\sqrt{25}\) OB= 5 cm In In \( \triangle O N D\) \( \mathrm{OD}=\mathrm{OB}=5 \mathrm{cm}\) \( \mathrm{ND}=\frac{\mathrm{CD}}{2}=\frac{8}{2}=4 \mathrm{cm}\) \( \mathrm{ON}^{2}+\mathrm{ND}^{2}=\mathrm{OD}^{2}\) \( \mathrm{ON}^{2}+(4)^{2}=(5)^{2}\) \( \mathrm{ON}^{2}=25-16=9\) ON = 3 cm Therefore, the distance of the bigger chord from the centre is 3 cm.

Q.4:Let the vertex of an angle ABC be located outside a circle and let the sides of the angle intersect equal chords AD and CE with the circle. Prove that ∠ABC is equal to half the difference of the angles subtended by the chords AC and DE at the centre.

Ans :In \( \triangle \mathrm{AOD}\) and \( \triangle \mathrm{COE}\) OA = OC (Radii Of the same circle) OD = OE (Radii of the same circle) AD = CE (Given) \( \angle A O D \angle C O E\) (SSS congruence rule ) \( \angle O A D=\angle O C E\) (by CPCT).............(1) \( \angle O D A=\angle O E C\) (by CPCT).............(2) Also, \( \angle O A D=\angle O D A(A S O A=O D)\)..............(3) From equations (1), (2), and (3), we obtain \( \angle O A D=\angle O C E=\angle O D A=\angle O E C\) Let \( \angle O A D=\angle O C E=\angle O D A=\angle O E C=x\) In \(\triangle \mathrm{O} \mathrm{A} \mathrm{C}\) OA=OC \( \angle O C A=\angle O A C\) (Let a) In \(\triangle \mathrm{O} \mathrm{D} \mathrm{E}\) OD=OE \( \angle O E D=\angle O D E\) (Let y) ADEC is a cyclic quadrilateral, \( \angle C A D+\angle D E C=180^{\circ}\) \( x+a+x+y=180^{\circ}\) \( 2 x+a+y=180^{\circ}\) \( y=180^{\circ}-2 x-a \ldots(4)\).................(4) However \( \angle A O C=180^{\circ}-2 a\) And \( \angle A O C=180^{\circ}-2 a\) \( \angle D O E-\angle A O C=2 a-2 y=2 a-2\left(180^{\circ}-2 x-a\right)\) \( =4 a+4 x-360^{\circ}\)....................(5) \( \angle B A C+\angle C A D=180^{\circ}\) \( \angle B A C=180^{\circ}-\angle C A D=180^{\circ}-(a+x)\) Similarly \( \angle A C B=180^{\circ}-(a+x)\) In \(\triangle \mathrm{A} \mathrm{B} \mathrm{C}\) \( \angle A B C+D B A C+D A C B=180^{\circ}\) (Angle sum property of a triangle) \( \angle A B C=180^{\circ}-\angle B A C-\angle A C B\) \( =180^{\circ}-\left(180^{\circ}-a-x\right)-\left(180^{\circ}-a-x\right)\) \( =2 a+2 x-180^{\circ}\) \( =\frac{1}{2}\left[4 a+4 x-360^{\circ}\right]\) \( \angle A B C=\frac{1}{2}[\angle D O E-\angle A O C][U \operatorname{sing} \text { equation }(5)]\)

Q.5:Prove that the circle drawn with any side of a rhombus as diameter, passes through the point of intersection of its diagonals.

Ans :Let } A B C D be a rhombus in which diagonals are intersecting at point O and a circle is drawn while taking side CD as its diameter. We know that a diameter subtends \(90^{\circ}\) on the arc. \( \angle C O D=90^{\circ}\) Also, in rhombus, the diagonals intersect each other at \(90^{\circ}\) \( \angle A O B=\angle B O C=\angle C O D=\angle D O A=90^{\circ}\) Clearly ,point O has to lie on the circle.

Q.6:ABCD is a parallelogram. The circle through A, B and C intersect CD (produced if necessary) at E. Prove that AE = AD.

Ans :It can be observed that ABCE is a cyclic quadrilateral and in a cyclic quadrilateral, the sum of the opposite angles is \(180^{\circ}\) \( \angle A E C+\angle C B A=180^{\circ}\) \( \angle A E C+\angle A E D=180^{\circ}\) \( \angle A E D=\angle C B A\)...............(1) For a parallelogram, opposite angles are equal. \( \angle A D E=\angle C B A \)......................(2) From (1) and (2) \( \angle A E D=\angle A D E\) AD = AE (Angles opposite to equal sides of a triangle)

Q.7:AC and BD are chords of a circle which bisect each other. Prove that (i) AC and BD are diameters, (ii) ABCD is a rectangle.

Ans :Let two choms Aa and CD are intersecting each other at point O. In\( \triangle A O B \text { and } \Delta C O D\), OA = OC (Given) OB = OD (Given) \( \angle A O B=\angle C O D\) \( \triangle \mathrm{AOB} \triangle \mathrm{COD}\) (SAS congruence rules) AB=CD (by CPCT) Similarly , it can be proved that \( \triangle A O D \triangle C O B\) AD= CB (by CPCT) Since in quadrilateral ACBD, opposite sides are equal in length, ACBD is a parallelogram. We know that opposite angles of a parallelogram are equal. OA=OC However OA+ OC=180 OA+OA=180 2 OA =180 OA =90 As ACBD is a parallelogram and one of its interior angles is 900, therefore, it is a rectangle. \(\angle A\) is the angle subtended by chord BD. And as \(\angle A=90^{\circ}\), therefore, B Dshould be the diameter of the circle. Similarly, AC is the diameter of the circle.

Q.8:Bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E and F respectively. Prove that the angles of the triangle DEF are 90° – 1/2 A, 90° – 1/2 B and 90° – 1/ 2 C.

Ans :It is given that 3t is the bisector of OB. \( \angle A B E=\frac{\angle B}{2}\) However \( \angle A D E=\angle A B E\) \( \angle A D E=\frac{\angle B}{2}\) Similarly \( \operatorname{DACF}=\angle A D F=\frac{\angle C}{2}\) \( \angle D=\angle A D E+\angle A D F\) \( =\frac{1}{2}(\angle B+\angle C)\) \( =\frac{1}{2}\left(180^{\circ}-\angle A\right)\) \( =90^{\circ}-\frac{1}{2} \angle \mathrm{A}\) Similarly it can be proved that \( \angle \mathrm{E}=90^{\circ}-\frac{1}{2} \angle \mathrm{B}\) \( \angle \mathrm{F}=90^{\circ}-\frac{1}{2} \angle \mathrm{C}\)

Q.9:Two congruent circles intersect each other at points A and B. Through A any line segment PAQ is drawn so that P, Q lie on the two circles. Prove that BP = BQ.

Ans :AB is the common chord in both the congruent circles. \( \angle A P B=\angle A Q B\) In \( \Delta B P Q\) \( \angle A P B=\angle A Q B\) \( \angle \mathrm{BQ}=\mathrm{BP}\){ Angles opposite to equal sides of a triangle) }

Q.10:In any triangle ABC, if the angle bisector of ∠A and perpendicular bisector of BC intersect, prove that they intersect on the circumcircle of the triangle ABC.

Ans :Let perpendicular bisector of side 3C and angle bisector of OA meet at point D. Let the perpendicular bisector of side 3C intersect it at E. Perpendicular bisector of side BC will pass through circumcentre O of the circle.\( \angle BOC\) and \( \angle BAC\) are the angles subtended by arc BC at the centre and a point A on the remaining part of the circle respectively. We also 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. \( \angle B O C=2 \angle B A C=2 \angle A\)........(1) In \( \Delta B O E \text { and } \Delta C O E\) OE=OE (common) OB=OC \( \angle O E B=\angle O E C\) \( \angle B O E \angle C O E\) However \( \angle B O E+\angle C O E=\angle B O C\) \( \angle B O E+\angle B O E=2 \angle A\) \( \angle 2 \angle B O E=2 \angle A\) \( \angle B O E=\angle A\) \( \angle B O E=\angle C O E=\angle A\) The perpendicular bisector Of side BC and angle bisector Of ÜA meet at point D. \( \angle B O D=\angle B O E=\angle A\)...........(3) Since AD is the bisector of angle OA \( \angle 2 \angle B A D=\angle A\)...........(4) From equations (3) and (4), we obtain \( \angle B O D=2 \angle B A D\) This can be possible only when point BD Will be a chord Of the circle. For this, the point D lies on the circum circle. Therefore, the perpendicular bisector of side BC and the angle bisector of OA meet on the circum circle of triangle ABC.

Q.1:How many tangents can a circle have?

Ans :Tangent is a line that intersects the circle at one point There are infinite number of points on circle At every point, there is one tangent Hence, there are infinite number of tangents in a circle.

Q.2:Fill in the blanks : (i) A tangent to a circle intersects it in point (s). (ii) A line intersecting a circle in two points is called a . (iii) A circle can have parallel tangents at the most. (iv) The common point of a tangent to a circle and the circle is called .

Ans :(i) Note only there can be one tangent at point P i.e. tangent XY If we try to make more than one line at point P example AB,it becomes a secant (as it intersects at more than one point) (ii) (iii) Note only two tangents are possible AB and XYCD is not a tangent it is a secant as it intersects at 2 points (iv) Here P is the point of contact.

Q.3:A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q so that OQ = 12 cm. Length PQ is : (A) 12 cm (B) 13 cm (C) 8.5 cm (D) \( \sqrt{119} \mathrm{cm}\)

Ans :Given OP radius 5 cm & OQ = 12 cm Since PQ is a tangent, \(\mathrm{OP} \perp \mathrm{PQ}\) \(\begin{array}{l}{\text { So, } \angle \mathrm{OPQ}=90^{\circ}} \\ {\text { Hence, } \Delta \mathrm{OPQ} \text { is a right triangle }}\end{array}\) \(\Delta \mathrm{OPQ}\) Using Pythagoras theorem \(\begin{array}{l}{\text { (Hypotenuse } )^{2}=(\text { Height })^{2}+(\text { Base })^{2}} \\ {\mathrm{OQ}^{2}=(\mathrm{OP})^{2}+(\mathrm{PQ})^{2}} \\ {12^{2}=5^{2}+(\mathrm{PQ})^{2}} \\ {144=25+(\mathrm{PQ})^{2}} \\ {144-25=\mathrm{PQ}^{2}} \\ {\mathrm{PQ}=\sqrt{119}}\end{array}\) So (D) is correct

Q.1:From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is (A) 7 cm (B) 12 cm (C) 15 cm (D) 24.5 cm

Ans :Let O be the centre of the circle. Given that, OQ=25cm and PQ=24 cm As the radius is perpendicular to the tangent at the point the contact Therefore ,\( OP \perp PQ\) Applying pythagoras theorem in triangle OPQ, we obtain \( \mathrm{OP}^{2}+\mathrm{PQ}^{2}=\mathrm{OQ}^{2}\) \( \mathrm{OP}^{2}+24^{2}=25^{2}\) \( \mathrm{OP}^{2}=625-576\) \( \mathrm{OP}^{2}=49\) OP=7 Therefore , the radius of circle is 7 cm. Hence ,alternative (A) is correct.

Q.2:In Fig. 10.11, if TP and TQ are the two tangents to a circle with centre O so that \( \angle \mathrm{PTQ}\)= \( 110^{\circ}\), then \( \angle \mathrm{PTQ}\) is equal to (A) \( 60^{\circ}\) (B) \( 70^{\circ}\) (C) \( 80^{\circ}\) (D) \( 90^{\circ}\)

Ans :It is given that TP and TQ are tangents. Therefore, radius drawn to these tangents will be perpendicular to the tangents. Thus, \(OP \perp TP\) and \(OQ \perp TO\) Angle OPT= \( 90^{\circ}\) Angle OQT = \( 90^{\circ}\) In quadrilateral POQT, Sum of all interior angles = \( 360^{\circ}\) \( \angle O P T+\angle P O Q+\angle O Q T+\angle P T Q=360^{\circ}\) \( 90^{\circ}+110^{\circ}+90^{\circ}+\angle P T Q=360^{\circ}\) \( \angle \mathrm{PTQ}=70^{\circ}\) Hence ,alternative (B) is correct.

Q.3:If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of \( 80^{\circ}\), then \( \angle \mathrm{POA}\) is equal to (A) \( 50^{\circ}\) (B) \( 60^{\circ}\) (C) \( 70^{\circ}\) (D) \( 80^{\circ}\)

Ans :It is given that PA and PB are tangents Therefore, the radius drawn to these tangents will be perpendicular to the tangents. Thus, \( OA \perp PA\) and \( 0B \perp PB\) \( \angle \mathrm{OBP}=90^{\circ}\) \( \angle \mathrm{OAP}=90^{\circ}\) In AOBP, Sum of all interior angles = \( 360^{\circ}\) \( \angle O A P+\angle A P B+\angle P B O+\angle B O A=360^{\circ}\) \( 90^{\circ}+80^{\circ}+90^{\circ}+\angle B O A=360^{\circ}\) \( \angle B O A=100^{\circ}\) \( \triangle O P B \text { and } \Delta O P A_{\prime}\) AP=BP( tangents from a point) OA=OB(radii of the circle) OP=OP(common side) Therefore ,\( \triangle O P B \approx \triangle O P A\) ( SSS congruence criterion) \( A \mapsto B, P \rightarrow P, O \rightarrow O\) And thus, \( \angle \mathrm{POB}=\angle \mathrm{POA}\) \( \angle \mathrm{POA}=\frac{1}{2} \angle \mathrm{AOB}=\frac{100^{\circ}}{2}=50^{\circ}\) Hence , alternative (A) is correct.

Q.4:Prove that the tangents drawn at the ends of a diameter of a circle are parallel.

Ans :Let AB be a diameter of the circle. Two tangents PQ and RS are drawn at points A and B respectively. Radius drawn to these tangents will be perpendiculæ to the tangents. Thus, \( OA \perp RS\) and \( 0B \perp PQ \) \( \angle O A R=90^{\circ}\) \( \angle O A S=90^{\circ}\) \( \angle O B P=90^{\circ}\) \( \angle O B Q=90^{\circ}\) I can be observed that , \( \angle O A R=\angle O B Q\) (Alternate interior angle) \( \angle O A S=\angle O B P\) (Alternate interior angle) Since alternate interior angles equal, lines PQ and RS will be parallel.

Q.5:Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.

Ans :Let us consider a circle With centre O. Let AB be a tangent which touches the circle at P. We have to prove that the line to AB at p passes through centre O. We shallprove this by contradiction method. Let us assume that the perpendicular to AB at P does not pass through centre O. Let it pass through another point O'. Join OP and O'P. As perpendicular to AB at P passes through O', therefore, \( \angle O^{\prime} P B=90^{\circ}\) ......(1) O is tie centre of the circle and P is the point of contact. We know the line joining the centre and the point Of contact to the tangent Of the circle are perpendicular to each other. \( \angle O P B=90^{\circ}\)................(2) Comparing equations (1) and (2), we obtain \( \angle O^{\prime} P B=\angle O P B\)..............(3) From the figure, it can be observed that, \( \angle O^{\prime} P B<\angle O P B\)...............(4) Therefore , \( \angle O^{\prime} P B<\angle O P B\) is not possible .it is only possible , when the line \( \mathrm{O}^{\prime} \mathrm{P}\) coincides with OP. Therefore the perpendicular to AB through P passes through centre O.

Q.6:The length of a tangent from a point A at distance 5 cm from the centre of the circle is 4 cm. Find the radius of the circle.

Ans :Let us consider a circle at point O. AB is a tangent drawn on this circle from point A. Given that, OA = 5cm and AB = 4 cm In triangle ABO, \( O B \perp A B\) Applying the pythagoras theorem in triangle ABO , we obtain \( A B^{2}+B O^{2}=O A^{2}\) \( 4^{2}+\mathrm{BO}^{2}=5^{2}\) \( 16+B O^{2}=25\) \( \mathrm{BO}^{2}=9\) BO=3 cm Hence , the radius of the circle is 3 cm

Q.7:Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.

Ans :Let the two concentric circles be centered at point O. And let PQ be the chord of the larger circle which touches the smaller circle at point A, Therefore, PQ is tangent to the smaller- circle, \( OA \perp PQ\) (As OA is the radius Of the circle) Applying Pythagoras theorem in triangle OAP, we obtain \( \mathrm{OA}^{2}+\mathrm{AP}^{2}=\mathrm{OP}^{2}\) \( 3^{2}+A P^{2}=5^{2}\) \( 9+A P^{2}=25\) \( \mathrm{AP}^{2}=16\) AP=4 Since \( OA\perp PQ \) AP = AQ (Perpendicular from the center Of the circle bisects the chord) \( \therefore P Q=2 A P=2 \times 4=8\) Therefore, the length of the chord of the larger circle is 8 cm.

Q.8:A quadrilateral ABCD is drawn to circumscribe a circle (see Fig. 10.12). Prove that AB + CD = AD + BC.

Ans :It can be observed that DR = DS (Tangents on the circle from point D) …………… (1) CR = CQ (Tangents on the circle from point C) …………… (2) BP = BQ (Tangents on the circle from point B) …………… (3) AP = AS on the circle from point A) ……………………… (4) Adding all these equations, we obtain \( \mathrm{DR}+\mathrm{CR}+\mathrm{BP}+\mathrm{AP}=\mathrm{DS}+\mathrm{CQ}+\mathrm{BQ}+\mathrm{AS}\) \( (DR+CR)+(BP+AP)=(DS+AS)+(CQ+BQ)\) \( CD+AB=AD+BC\)

Q.9:In Fig. 10.13, XY and \( X^{\prime} Y^{\prime}\) are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and \(X^{\prime} Y^{\prime}\) at B. Prove that \( \angle \mathrm{AOB}\)= \( 90^{\circ}\).

Ans :Let us joint point O to C. \( \triangle \mathrm{OPA} \text { and } \triangle \mathrm{OCA}_{\mathrm{r}}\) OP = OC (Radii of the same circle) AP=AC (Tangents from point A) AO = AO (Common side) \( \Delta O P A \cong \triangle O C A\) (SSS congruence criterion) Therefore, \( \mathrm{P} \mapsto \mathrm{C}, \mathrm{A} \mapsto \mathrm{A}, \mathrm{O} \mapsto \mathrm{O}\) \( \angle \mathrm{POA}=\angle \mathrm{COA}\)..................(1) Similarly \( \triangle O Q B \cong \triangle O C B\) \( \angle Q O B = ( \angle C O S \).......................(2) Since P O Q is a diameter of the circle, it is a straight line. Therefore \( \angle P O A+\angle C O A+\angle C O B+\angle Q O B=180^{\circ}\) From equations (1) and (2), it can be observed that \( 2 \angle C O A+2 \angle C O B=180^{\circ}\) \( \angle \mathrm{COA}+\angle \mathrm{COB}=90^{\circ}\) \( \angle A O B=90^{\circ}\)

Q.10:Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.

Ans :Let us consider a circle centered at point O. Let P be external point from which two tangents PA and PB are drawn to the circle which are touching the circle at point A and B respectively and AB is the line segment, joining point Of contacts A and 3 together such that it subtends Angle AOB at centre- O of the circle. It can be observed that \( OA (radius) \perp PA (tangent)\) Therefore ,\( \angle O A P=90^{\circ}\) Similarly , \(OB \perp PB (tangent)\) \( \angle \mathrm{OBP}=90^{\circ}\) In quadrilateral OAPB, Sum of the all interior angles \( \angle O A P+\angle A P B+\angle P B O+\angle B O A=360^{\circ}\) \( 900+\angle A P B+90^{\circ}+\angle B O A=360^{\circ}\) \( \angle A P B+\angle B O A=180^{\circ}\) Hence, it can be observed that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line- segment joining the points of contact at the centre.

Q.11:Prove that the parallelogram circumscribing a circle is a rhombus.

Ans :Since ABCO is a parallelogram, AB=AD…………(1) BC=AD…………(2) It can be observed that DR = DS (Tangents on the circle from point D) CR = CQ (Tangents on the circle from point C) BP=BQ (Tangents on the circle from point B) AP= AS (Tangents on the circle from point A) Adding all these equations, we obtain OR + CR + BP + AP DS + CQ + BQ + AS (DR + CR) + (BP + AD) = (DS + AS) + (CQ + BQ) CD + AB = AD + BC On putting the values of equations (1 ) and (2) in this equation, we obtain 2A8 =2BC AB=BC……….(3) Comparing equations (1), (2), (3), we obtain AS = BC = CD = DA Hence, ABCD is a rhombus.

Q.12:A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively (see Fig. 10.14). Find the sides AB and AC.

Ans :Let the given circle touch the sides AB and AC of the triangle at point E and F respectively and the Length of the line segment AF be x. In triangle ABC, CF=CD =6cm (Tangents on the circle from point C) BE= BD = 8cm (Tangents on the circle from point B) AE=AF =x cm (Tangents on the circle from point A) AB=AE+EB=x + 8 BC=BD+ DC = 8+6=14 CA=CF+FA=6+x 2s=AB + BC + CA =x+8+14+6+x =28+2x s=14+x \( \Delta \mathrm{ABC}=\sqrt{s(s-a)(s-b)(s-c)}\) \( =\sqrt{\{14+x\}\{(14+x)-14\}\{(14+x)-(6+x)\}(14+x)-(8+x)\}}\) \( =\sqrt{(14+x)(x)(8)(6)}\) \( =4 \sqrt{3\left(14 x+x^{2}\right)}\) Area of \( \Delta \mathrm{OBC}=\frac{1}{2} \times \mathrm{OD} \times \mathrm{BC}=\frac{1}{2} \times 4 \times 14=28\) Area of \( \Delta \mathrm{OCA}=\frac{1}{2} \times \mathrm{OF} \times \mathrm{AC}=\frac{1}{2} \times 4 \times(6+x)=12+2 x\) Area of \( \frac{1}{2} \times \mathrm{OE} \times \mathrm{AB}=\frac{1}{2} \times 4 \times(8+x)=16+2 x\) \( \Delta \mathrm{ABC}=\text { Area of } \Delta \mathrm{OBC}+\text { Area of } \Delta \mathrm{OCA}+\text { Area of } \triangle \mathrm{OAB}\) \( 4 \sqrt{3\left(14 x+x^{2}\right)}=28+12+2 x+16+2 x\) \( \Rightarrow 4 \sqrt{3\left(14 x+x^{2}\right)}=56+4 x\) \( \Rightarrow \sqrt{3\left(14 x+x^{2}\right)}=14+x\) \( \Rightarrow 3\left(14 x+x^{2}\right)=(14+x)^{2}\) \( \Rightarrow 42 x+3 x^{2}=196+x^{2}+28 x\) \( \Rightarrow 2 x^{2}+14 x-196=0\) \( \Rightarrow x^{2}+7 x-98=0\) \( \Rightarrow x^{2}+14 x-7 x-98=0\) \( \Rightarrow x(x+14)-7(x+14)=0\) \( \Rightarrow(x+14)(x-7)=0\) Either x+14=0 or x-7=0 Therefore ,x = -1 and 7 However, x = -14 is not possible as the length of the sides will be negative. Therefore, x=7 Hence, AB =x+8 = 7+8=15 cm CA = 6 + x=6 + 7 =13 cm

Q.13:Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

Ans :Let ABCD be a quadrilateral circumscribing a circle centered at O such that it touches the circle at point P, Q, R, S. Let us join the v«tices of the quadrilateral ABCD to the of the circle. Consider triangle OAP and triangle OAS, AP = AS (Tangents from the sarne point) OP=OS (Radii Of the sane circle) OA=OA (Common side) \( \triangle O A P=\triangle O A S\) ( SSS congruence criterion) Therefore ,A <-> A,P <-> S ,O <-> O And thus \( \angle \mathrm{POA}=\angle \mathrm{AOS}\) \( \angle 1=\angle 8\), Similarly, \( \angle 2=\angle 3\), \( \angle 4=\angle 5\), \( \angle 6=\angle 7\), \( \angle 1+\angle 2+\angle 3+\angle 4+\angle 5+\angle 6+\angle 7+\angle 8=360^{\circ}\) \( (\angle 1+\angle 8)+(\angle 2+\angle 3)+(\angle 4+\angle 5)+(\angle 6+\angle 7)=360^{\circ}\) \( 2 \angle 1+2 \angle 2+2 \angle 5+2 \angle 6=360^{\circ}\) \( 2(\angle 1+\angle 2)+2(\angle 5+\angle 6)=360^{\circ}\) \( (\angle 1+\angle 2)+(\angle 5+\angle 6)=180^{\circ}\) Similarly , we can prove that \( \angle B O C+ \angle D O A=180^{\circ}\) Hence, opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle .

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