## NCERT Solutions for Class 9 Maths Chapter 10 Circles Ex 10.6

**The topics and sub-topics in NCERT Class 9 Maths Textbook Chapter 10 Circles:**

- Circles
- Introduction
- Circles And Its Related Terms: A Review
- Angle Subtended By A Chord At A Point
- Perpendicular From The Centre To A Chord
- Circle Through Three Points
- Equal Chords And Their Distances From The Centre
- Angle Subtended By An Arc Of A Circle
- Cyclic Quadrilaterals
- Summary

Formulae Handbook for Class 9 Maths and Science Educational Loans in India

- Class 9 Maths Chapter 10 Circles NCERT Solutions Ex 10.1
- Class 9 Maths Chapter 10 Circles NCERT Solutions Ex 10.2
- Class 9 Maths Chapter 10 Circles NCERT Solutions Ex 10.3
- Class 9 Maths Chapter 10 Circles NCERT Solutions Ex 10.4
- Class 9 Maths Chapter 10 Circles NCERT Solutions Ex 10.5
- Extra Questions for Circles

### NCERT Solutions for Class 9 Maths Chapter 10 Circles Ex 10.6

**Ex 10.6 Class 9 Maths Question 1.**

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

**Solution:**

**Given:** Two circles with centres O and O’ which intersect each other at C and D.

**To prove:** ∠OCO’ = ∠ODO’

**Construction:** Join OC, OD, O’C and O’D

**Proof:** In ∆ OCO’and ∆ODO’, we have

OC = OD (Radii of the same circle)

O’C = O’D (Radii of the same circle)

OO’ = OO’ (Common)

∴ By SSS criterion, we get

∆ OCO’ ≅ ∆ ODO’

Hence, ∠OCO’ = ∠ODO’ (By CPCT)

**Ex 10.6 Class 9 Maths Question 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.**

**Solution:**

Let O be the centre of the given circle and let its radius be cm.

Draw ON ⊥ AB and OM⊥ CD since, ON ⊥ AB, OM ⊥ CD and AB || CD, therefore points N, O, M are collinear.

Let ON = a cm

∴ OM = (6 – a) cm

Join OA and OC.

Then, OA = OC = b c m

Since, the perpendicular from the centre to a chord of the circle bisects the chord.

Therefore, AN = NB= 2.5 cm and OM = MD = 5.5 cm

In ∆OAN and ∆OCM, we get

OA^{2} = ON^{2} + AN^{2}

OC^{2} = OM^{2} + CM^{2}

⇒ b^{2} = a^{2} + (2.5)^{2}

and, b^{2} = (6-a)^{2} + (5.5)^{2} …(i)

So, a^{2} + (2.5)^{2} = (6 – a)^{2} + (5.5)^{2}

⇒ a^{2} + 6.25= 36-12a + a^{2} + 30.25

⇒ 12a = 60

⇒ a = 5

On putting a = 5 in Eq. (i), we get

b^{2} = (5)^{2} + (2.5)^{2}

= 25 + 6.25 = 31.25

So, r = \( \sqrt{31.25} \) = 5.6cm (Approx.)

**Ex 10.6 Class 9 Maths Question 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 ?**

**Solution:**

Let PQ and RS be two parallel chords of a circle with centre O such that PQ = 6 cm and RS = 8 cm.

Let a be the radius of circle.

Draw ON ⊥ RS, OM ⊥ PQ. Since, PQ || RS and ON ⊥ RS, OM⊥ PQ, therefore points 0,N,M are collinear.

∵ OM = 4 cm and M and N are the mid-points of PQ and RS respectively.

PM = MQ = \(\frac { 1 }{ 2 }\) PQ = \(\frac { 6 }{ 2 }\) = 3 cm

and RN = NS = \(\frac { 1 }{ 2 }\) RS = \(\frac { 8 }{ 2 }\) = 4 cm

In ∆OPM, we have

OP^{2} = OM^{2} + PM^{2}

⇒ a^{2} =4^{2} + 3^{2} = 16 + 9 = 25

⇒ a = 5

In ∆ORN, we have

⇒ OR^{2} = ON^{2} + RN^{2}

⇒ a^{2} = ON^{2} + (4)^{2}

⇒ 25 = ON^{2} + 16

⇒ ON^{2} = 9

⇒ ON = 3cm

Hence, the distance of the chord PS from the centre is 3 cm.

**Ex 10.6 Class 9 Maths Question 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.**

**Solution:**

Since, an exterior angle of a triangle is equal to the sum of the interior opposite angles.

∴ In ∆BDC, we get

∠ADC = ∠DBC + ∠DCB …(i)

Since, angle at the centre is twice at a point on the remaining part of circle.

∴ ∠DCE = \(\frac { 1 }{ 2 }\) ∠DOE

⇒ ∠DCB = \(\frac { 1 }{ 2 }\) ∠DOE (∵ ∠DCE = ∠DCB)

∠ADC = \(\frac { 1 }{ 2 }\) ∠AOC

∴ \(\frac { 1 }{ 2 }\) ∠AOC = ∠ABC + \(\frac { 1 }{ 2 }\) ∠DOE (∵ ∠DBC = ∠ABC)

∴ ∠ABC = \(\frac { 1 }{ 2 }\) (∠AOC – ∠DOE)

Hence, ∠ABC is equal to half the difference of angles subtended by the chords AC and DE at the centre.

**Ex 10.6 Class 9 Maths Question 5.**

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

**Solution:**

**Given:** PQRS is a rhombus. PR and SQ are its two diagonals which bisect each other at right angles.

**To prove:** A circle drawn on PQ as diameter will pass through O.

**Construction:** Through O, draw MN || PS and EF || PQ.

**Proof :** ∵ PQ = SR ⇒ \(\frac { 1 }{ 2 }\) PQ = \(\frac { 1 }{ 2 }\) SR

So, PN = SM

Similarly, PE = ON

So, PN = ON = NQ

Therefore, a circle drawn with N as centre and radius PN passes through P, O, Q.

**Ex 10.6 Class 9 Maths Question 6.**

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

**Solution:**

Since, ABCE is a cyclic quadrilateral, therefore

∠AED+ ∠ABC= 180°

(∵ Sum of opposite angle of a cyclic quadrilateral is 180°) .. .(i)

∵ ∠ADE + ∠ADC = 180° (EDC is a straight line)

So, ∠ADE + ∠ABC = 180°

(∵ ∠ADC = ∠ABC opposite angle of a || gm).. .(ii)

From Eqs. (i) and (ii), we get

∠AED + ∠ABC = ∠ADE + ∠ABC

⇒ ∠AED = ∠ADE

∴ In ∆AED We have

∠AED = ∠ADE

So, AD = AE

(∵ Sides opposite to equal angles of a triangle are equal)

**Ex 10.6 Class 9 Maths Question 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.**

**Solution:**

(i) Let BD and AC be two chords of a circle bisect at P.

In ∆APB and ∆CPD, we get

PA = PC ( ∵ P is the mid-point of AC)

∠APB = ∠CPD (Vertically opposite angles)

and PB = PD (∵ P is the mid-point of BD)

∴ By SAS criterion

∆CPD ≅ ∆APB

∴ CD= AB (By CPCT) …(i)

∴ BD divides the circle into two equal parts. So, BD is a diameter.

Similarly, AC is a diameter.

(ii) Now, BD and AC bisect each other.

So, ABCD is a parallelogram.

Also, AC = BD

∴ ABCD is a rectangle.

**Ex 10.6 Class 9 Maths Question 8.**

**Bisectors of angles A, B and C of a ∆ABC intersect its circumcircle at D, E and F, respectively. Prove that the angles of the ∆DEF are 90° – \(\frac { 1 }{ 2 }\) A, 90° – \(\frac { 1 }{ 2 }\) B and 90° – \(\frac { 1 }{ 2 }\) C.**

**Solution:**

∵ ∠EDF = ∠EDA + ∠ADF

∵ ∠EDA and ∠EBA are the angles in the same segment of the circle.

∴ ∠EDA = ∠EBA

and similarly ∠ADF and ∠FCA are the angles in the same segment and hence

**Ex 10.6 Class 9 Maths Question 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.**

**Solution:**

Let O’ and O be the centres of two congruent circles.

Since, AB is a common chord of these circles.

∴ ∠BPA = ∠BQA

(∵ Angle subtended by equal chords are equal)

⇒ BP = BQ

**Ex 10.6 Class 9 Maths Question 10.**

**In any ∆ ABC, if the angle bisector of ∠A and perpendicular bisector of BC intersect, prove that they intersect on the circumcircle of the ∆ABC.**

**Solution:**

(i) Let bisector of ∠A meet the circumcircle of ∆ABC at M.

Join BM and CM.

∴ ∠MBC = ∠MAC (Angles in same segment)

and ∠BCM = ∠BAM (Angles in same segment)

But ∠BAM = ∠CAM (∵ AM is bisector of ∠A)…. .(i)

∴ ∠MBC = ∠BCM

So, MB = MC (Sides opposite to equal angles are equal)

So, M must lie on the perpendicular bisector of BC

(ii) Let M be a point on the perpendicular bisector of BC which lies on circumcircle of ∆ ABC.

Join AM.

Since, M lies on perpendicular bisector of BC.

∴ BM = CM

∠MBC = ∠MCB

But ∠MBC = ∠MAC (Angles in same segment)

and ∠MCB = ∠BAM (Angles in same segment)

So, from Eq. (i),

∠BAM = ∠CAM

AM is the bisector of A.

Hence, bisector of ∠A and perpendicular bisector of BC at M which lies on circumcircle of ∆ABC.

### NCERT Solutions for Class 9 Maths Chapter 10 Circles (वृत्त) (Hindi Medium) Ex 10.6

## NCERT Solutions for Class 9 Maths

- Chapter 1 Number systems
- Chapter 2 Polynomials
- Chapter 3 Coordinate Geometry
- Chapter 4 Linear Equations in Two Variables
- Chapter 5 Introduction to Euclid Geometry
- Chapter 6 Lines and Angles
- Chapter 7 Triangles
- Chapter 8 Quadrilaterals
- Chapter 9 Areas of Parallelograms and Triangles
- Chapter 10 Circles
- Chapter 11 Constructions
- Chapter 12 Heron’s Formula
- Chapter 13 Surface Areas and Volumes
- Chapter 14 Statistics
- Chapter 15 Probability
- Class 9 Maths (Download PDF)