# Class 9 NCERT Solutions Maths Chapter 10 - Circles

Score marks in short answer questions easily by practising with our NCERT Solutions for CBSE Class 9 Mathematics Chapter 10 Circles. Write and practise with TopperLearning’s expert solutions for long answer questions based on chords. By revising a variety of textbook questions, you’ll learn how to prove chord properties in a step-wise format.

With CBSE Class 9 Maths chapter solutions, learn the application of important concepts such as RHS congruence rule, angle sum property of a triangle etc. This knowledge will help you think logically and arrive at correct answers in your Maths exam. Other than expert solutions, you may want to use sample papers, online mock tests, concept videos etc. to revise circles for exam preparation.

## Circles Exercise Ex. 10.1

### Solution 1

(ii) A point, whose distance from the centre of a circle is greater than its radius lies in exterior of

(iii) The longest chord of a circle is a diameter of the circle.

(iv) An arc is a semicircle when its ends are the ends of a diameter.

(v) Segment of a circle is the region between an arc and chord of the circle.

(vi) A circle divides the plane, on which it lies, in three parts.

### Solution 2

## Circles Exercise Ex. 10.2

### Solution 1

So, if we try to superimpose two circles of equal radius, one each other both circles will cover each other.

So, two circles are congruent if they have equal radius.

AB = CD (chords of same length)

OA = O'C (radii of congruent circles)

OB = O'D (radii of congruent circles)

AOB CO'D (SSS congruence rule)

AOB = CO'D (by CPCT)

Hence equal chords of congruent circles subtend equal angles at their centres.

### Solution 2

AOB = CO'D (given)

OA = O'C (radii of congruent circles)

OB = O'D (radii of congruent circles)

AOB CO'D (SSS congruence rule)

AB = CD (by CPCT)

Hence, if chords of congruent circles subtend equal angles at their centres then chords are equal.

## Circles Exercise Ex. 10.3

### Solution 1

(i) circles don't intersect each other at any point, so circles are not having any point in common.

(iv) These circles intersect each other at two points P and Q. So the circles have two points in common. We may observe that there can be maximum 2 points in common.

We can have a situation in which two congruent circles are superimposed on each other, this situation can be referred as if we are drawing circle two times.

### Solution 2

Step1. Take the given circle centered at point O.

Step2. Take any two different chords AB and CD of this circle and draw perpendicular bisectors of these

Step3. Let these perpendicular bisectors meet at point O. Now, O is the centre of given circle.

### Solution 3

Join AB. AB is the chord for circle centered at O, so perpendicular bisector of AB will pass through O.

Again AB is also chord of circle centered at O', so, perpendicular bisector of AB will also pass through O'.

Clearly centres of these circles lie on the perpendicular bisector of common chord.

## Circles Exercise Ex. 10.4

### Solution 1

OA = OB = 5 cm

O'A = O'B = 3 cm

OO' will be the perpendicular bisector of chord AB.

AC = CB

Given that OO' = 4 cm

Let OC be x. so, O'C will be 4 - x

In OAC

OA

^{2}= AC

^{2}+ OC

^{2}

5

^{2}= AC

^{2}+ x

^{2}

25 - x

^{2}= AC

^{2}... (1)

In O'AC

O'A

^{2}= AC

^{2}+ O'C

^{2}

3

^{2}= AC

^{2}+ (4 - x)

^{2}

9 = AC

^{2}+ 16 + x

^{2}- 8x

AC

^{2}= - x

^{2}- 7 + 8x ... (2)

From equations (1) and (2), we have

25 - x

^{2}= - x

^{2}- 7 + 8x

8x = 32

x = 4

So, the common chord will pass through the centre of smaller circle i.e. O'. and hence it will be diameter of smaller circle.

^{2}= 25 - x

^{2}= 25 - 4

^{2}= 25 - 16 = 9

AC = 3 m

The length of the common chord AB = 2 AC = (2 3) m = 6 m

### Solution 2

In OVT and OUT

OV = OU (Equal chords of a circle are equidistant from the centre)

OVT = OUT (Each 90

^{o})

OT = OT (common)

It is given that

PQ = RS ... ... ... ... (2)

On adding equations (1) and (3), we have

PV + VT = RU + UT

PT = RT ... ... ... ... (4)

On subtracting equation (4) from equation (2), we have

PQ - PT = RS - RT

QT = ST ... ... ... ... (5)

Equations (4) and (5) shows that the corresponding segments of

chords PQ and RS are congruent to each other.

### Solution 3

Draw perpendiculars OV and OU on these chords.

In OVT and OUT

OV = OU (Equal chords of a circle are equidistant from the centre)

OVT = OUT (Each 90

^{o})

OT = OT (common)

OVT OUT (RHS congruence rule)

OTV = OTU (by CPCT)

### Solution 4

Here, BC is chord of smaller circle and AD is chord of bigger circle.

We know that the perpendicular drawn from centre of circle bisects the chord.

BM = MC ... (1)

And AM = MD ... (2)

Subtracting equations (2) from (1), we have

AM - BM = MD - MC

AB = CD

### Solution 5

Let R, S and M be the position of Reshma, Salma and Mandip respectively.

In OAR

OA

^{2}+ AR

^{2}= OR

^{2}

OA

^{2}+ (3 m)

^{2}= (5 m)

^{2}

OA

^{2}= (25 - 9) m

^{2}= 16 m

^{2}

OA = 4 m

We know that in an isosceles triangle altitude divides the base, so in RSM

RCS will be of 90

^{o}and RC = CM

Area of ORS = OARS

RC = 4.8

### Solution 6

So, ASD is a equilateral triangle

OA (radius) = 20 m.

Medians of equilateral triangle pass through the circum centre (O) of the equilateral triangle ABC.

We also know that median intersect each other at the 2: 1. As AB is the median of equilateral triangle ABC, we can write

In ABD

AD

^{2}= AB

^{2}+ BD

^{2}

AD

^{2}= (30)

^{2}+

So, length of string of each phone will be m.

## Circles Exercise Ex. 10.5

### Solution 1

AOC = AOB + BOC

= 60

^{o}+ 30

^{o}

= 90

^{o}

We know that angle subtended by an arc at centre is double the angle subtended by it any point on the remaining part of the circle.

### Solution 2

AB = OA = OB = radius

OAB is an equilateral triangle.

So, each interior angle of this triangle will be of 60

^{o}

^{o}

ACB + ADB = 180

^{o}(Opposite angle in cyclic quadrilateral)

ADB = 180

^{o}- 30

^{o}= 150

^{o}

So, angle subtended by this chord at a point on major arc and minor arc are 30

^{o}and 150

^{o}respectively.

### Solution 3

Consider PR as a chord of circle.

Take any point S on major arc of circle.

Now PQRS is a cyclic quadrilateral.

^{o}(Opposite angles of cyclic quadrilateral)

PSR = 180

^{o}- 100

^{o}= 80

^{o}

We know that angle subtended by an arc at centre is double the angle subtended by it any point on the remaining part of the circle.

POR = 2PSR = 2 (80

^{o}) = 160

^{o}

OP = OR (radii of same circle)

^{o}(Angle sum property of a triangle)

^{o}= 180

^{o}

2 OPR = 180

^{o}- 160

^{o}= 20

^{o}

^{o}

### Solution 4

^{o}(Angle sum property of a triangle)

BAC + 69

^{o}+ 31

^{o}= 180

^{o}

^{o}- 100º

BAC = 80

^{o}

^{o}(Angles in same segment of circle are equal)

### Solution 5

CDE + DCE = CEB (Exterior angle)

CDE + 20

^{o}= 130

^{o}

^{o}

But BAC = CDE (Angles in same segment of circle)

BAC = 110

^{o}

### Solution 6

^{o}

^{o}+ 70

^{o}= 100

^{o}

BCD + BAD = 180

^{o}(Opposite angles of a cyclic quadrilateral)

BCD + 100

^{o}= 180

^{o}

BCD = 80

^{o}

In ABC

BCA = CAB (Angles opposite to equal sides of a triangle)

BCA = 30

^{o}

We have BCD = 80

^{o}

BCA + ACD = 80

^{o}

30

^{o}+ ACD = 80

^{o}

^{o}

^{o}

### Solution 7

^{o}(Cyclic quadrilateral)

BCD = 180

^{o}- 90

^{o}= 90

^{o}

^{o}(Cyclic quadrilateral)

90

^{o}+ ABC = 180

^{o}

ABC = 90

^{o}

Here, each interior angle of cyclic quadrilateral is of 90

^{o}. Hence it is a rectangle.

### Solution 8

In AMD and BNC

AD = BC (Given)

AMD = BNC (By construction each is 90

^{o})

AM = BM (Perpendicular distance between two parallel lines is same)

AMD BNC (RHS congruence rule)

^{o}... (2)

BAD + BCD = 180

^{o}[Using equation (1)]

This equation shows that the opposite angles are supplementary.

So, ABCD is a cyclic quadrilateral.

### Solution 9

For chord AP

PBA = ACP (Angles in same segment) ... (1)

For chord DQ

DBQ = QCD (Angles in same segment) ... (2)

ABD and PBQ are line segments intersecting at B.

PBA = DBQ (Vertically opposite angles) ... (3)

From equations (1), (2) and (3), we have

ACP = QCD

### Solution 10

Two circles are drawn while taking AB and AC as diameter.

Let they intersect each other at D and let D does not lie on BC.

Join AD

ADB = 90

^{o}(Angle subtend by semicircle)

ADC = 90

^{o}(Angle subtend by semicircle)

BDC = ADB + ADC = 90

^{o}+ 90

^{o}= 180

^{o}

Hence BDC is straight line and our assumption was wrong.

Thus, Point D lies on third side BC of ABC

### Solution 11

^{o}(Angle sum property of a triangle)

90

^{o}+ BCA + CAB = 180

^{o}

BCA + CAB = 90

^{o}... (1)

In ADC

^{o}(Angle sum property of a triangle)

90

^{o}+ ACD + DAC = 180

^{o}

ACD + DAC = 90

^{o}... (2)

Adding equations (1) and (2), we have

^{o}

^{o}BCD + DAB = 180

^{o}... (3)

But it is given that

B + D = 90

^{o}+ 90

^{o}= 180

^{o}... (4)

From equations (3) and (4), we can see that quadrilateral ABCD is having sum of measures of opposite angles as 180

^{o}.

So, it is a cyclic quadrilateral.

Consider chord CD.

Now, CAD = CBD (Angles in same segment)

### Solution 12

A + C = 180

^{o}(Opposite angle of cyclic quadrilateral) ... (1)

We know that opposite angles of a parallelogram are equal

A = C and B = D

From equation (1)

A + C = 180

^{o}

A + A = 180

^{o}

2 A = 180

^{o}

A = 90

^{o}

Parallelogram ABCD is having its one of interior angles as 90

^{o}, so, it is a rectangle.

## Circles Exercise Ex. 10.6

### Solution 1

Construction: Let us join OO',

OA = OB (radius of circle 1)

O'A = O'B (radius of circle 2)

OO' = OO' (common)

OAO' = OBO' (by CPCT)

So, line of centres of two intersecting circles subtends equal angles at the two points of intersection.

### Solution 2

In MOB

So, from equation (1) and (2)

### Solution 3

OM = 4 cm

### Solution 4

OA = OC (radii of same circle)

OD = OE (radii of same circle)

AD = CE (given)

AOD COE (SSS congruence rule)

ODA = OEC (by CPCT) ... (2)

We also have

OAD = ODA (As OA = OD) ... (3)

From equations (1), (2) and (3), we have

OAD = OCE = ODA = OEC

Let OAD = OCE = ODA = OEC = x

In OAC,

OA = OC

OCA = OAC (let a)

OD = OE

OED = ODE (let y)

ADEC is a cyclic quadrilateral

CAD + DEC = 180

^{o}(opposite angles are supplementary)

^{o}

2x + a + y = 180

^{o}

y = 180 - 2x - a ... (4)

But DOE = 180 - 2y

And AOC = 180 - 2a

Now, DOE - AOC = 2a - 2y = 2a - 2 (180 - 2x - a)

= 4a + 4x - 360

^{o}... (5)

Now, BAC + CAD = 180 (Linear pair)

BAC = 180 - CAD = 180 - (a + x)

Similarly, ACB = 180 - (a + x)

Now, in ABC

ABC + BAC + ACB = 180 (Angle sum property of a triangle)

ABC = 180 - BAC - ACB

= 180 - (180 - a - x) - (180 - a -x)

= 2a + 2x - 180

= [4a + 4x - 360

^{o}]

ABC = [DOE - AOC] [Using equation (5)]

### Solution 5

We know that angle in a semicircle is of 90

^{o}.

COD = 90

^{o}

Also in rhombus the diagonals intersect each other at 90

^{o}

AOB = BOC = COD = DOA = 90

^{o}

So, point O has to lie on the circle.

### Solution 6

^{o}

AEC + CBA = 180

^{o}

AEC + AED = 180

^{o}(linear pair)

AED = CBA ... (1)

For a parallelogram opposite angles are equal.

ADE = CBA ... (2)

From (1) and (2)

AED = ADE

AD = AE (angles opposite to equal sides of a triangle)

### Solution 7

In AOB and COD

OA = OC (given)

OB = OD (given)

AOB = COD (vertically opposite angles)

AOB COD (SAS congruence rule)

AB = CD (by CPCT)

Similarly, we can prove AOD COB

AD = CB (by CPCT)

Hence, ACBD is a parallelogram.

We know that opposite angles of a parallelogram are equal

A = C

^{o}(ABCD is a cyclic quadrilateral)

A + A = 180

^{o}

A = 180

^{o}

A = 90

^{o}

As ACBD is a parallelogram and one of its interior angles is 90

^{o}, so it is a rectangle.

A is the angle subtended by chord BD. And as A = 90

^{o}, so BD should be diameter of circle. Similarly AC is diameter of circle.

### Solution 8

It is given that BE is the bisector of B

ABE =

But ADE = ABE (angles in same segment for chord AE)

ADE =

Similarly, ACF = ADF = (angle in same segment for chord AF)

Now, D = ADE + ADF

### Solution 9

APB = AQB

Now in BPQ

APB = AQB

BP = BQ (angles opposite to equal sides of a triangle)

### Solution 10

Let perpendicular bisector of side BC intersects it at E.

Perpendicular bisector of side BC will pass through circum centre O of circle. Now, BOC and 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.

BOC = 2 BAC = 2A ... (1)

In BOE and COE

OB = OC (radii of same circle)

^{o}as OD BC)

But BOE + COE = BOC

BOE = 2A

2BAD = A ... (4)

From equations (3) and (4), we have

BOD = 2 BAD

It is possible only if BD will be a chord of the circle. For this the point D lies on circum circle.

Therefore, the perpendicular bisector of side BC and angle bisector of A meet on the circum circle of triangle ABC.

## CBSE Class 9 Frequently Asked Questions

NCERT Solutions for Class 9 Maths Chapter 10 is on Circles and is a crucial chapter as prescribed in the Class 9 syllabus by NCERT. The chapter Circle of Class 9 Maths includes an introduction to circles, the angle subtended by the chord, perpendicular from centre to chord, properties of equal chords and cyclic quadrilateral. There are five exercises in the Class 9 Maths Chapter 10 on circles. Apart from the NCERT Solutions drafted by experienced and qualified experts, students are recommended to go through the Sample papers, chapter notes, and question bank of the chapter, Circle.

TopperLearning's doubt solver "Ask a doubt" feature quickly assists all your queries.

We should always follow a national curricular or framework drafted by eminent experts and follow NCERT Solutions for Class 9 Maths Chapter 10. The type of questions included across the five exercises of the chapter provides good exposure and experience to deal with different types of questions that can be asked in the school exams. Practising from NCERT Solutions for Class 9 Maths Chapter 10 will enable students to practice and attempt similar questions that may be asked in various exams. Once students are clear on the conceptual part and aware of the pattern of problems asked from a specific chapter, it becomes handy to tackle challenging problems based on that concept. It is also evident that a new typology of questions is included in the examinations like Case-based questions.

Students can get any time access to NCERT Solutions for Class 9 Maths Chapter 10, Circles on the TopperLearning website, where they can practice all the five exercises. Eminent and qualified experts carefully draft the solutions to these problems.

To understand Circles, a student must know about chords, diameter, radius, bisector and much more. The chapter Circle of Class 9 Maths includes an introduction to circles, the angle subtended by the chord, perpendicular from centre to chord, properties of equal chords and cyclic quadrilateral. There are five exercises in the Class 9 Maths Chapter 10 on circles, giving students an idea of different types of question patterns required to handle any questions in the exam.

To solve the problems of Class 9 Maths Chapter 10 Circles, one must understand the concepts first. With TopperLearning resources like the concept and problem-solving application videos, chapter-wise notes and question banks, students can clarify the concepts and then try to solve the NCERT solution independently.

The critical theorems of Class 9 Chapter 10 Circles are listed below.

- Theorem: Equal chords of a circle subtend equal angles at the centre.
- Theorem: This is the converse of the previous theorem. It implies that if two chords subtend equal angles at the centre, they are equal.
- Theorem: Equal chords of a circle are equidistant from the centre of a circle.
- Theorem: This is the converse of the previous theorem. It states that chords equidistant from the circle's centre are equal in length.
- Angle subtended by an arc at the centre of a circle is double that of an angle that arc subtends at any given point on the circle.
- Theorem: Angles formed in the same circle segment are always equal in measure.

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