# Class 9 NCERT Solutions Maths Chapter 9: Circles

Ex. 9.1

Ex. 9.2

Ex. 9.3

## Circles Exercise Ex. 9.1

### Solution 1

A circle is a collection of points which are equidistant from a fix point. This fix point is called as the centre of circle and this equal distance is called as radius of circle. And thus shape of a circle depends on the radius of the circle.

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.

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.

Now consider two congruent circles having centre O and O' and two chords AB and CD of equal lengths

Now in AOB and CO'D

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)

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

Let us consider two congruent circles (circles of same radius) with centres as O and O'.

In AOB and CO'D

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.

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. 9.2

### Solution 1

Let radius of circle centered at O and O' be 5 cm and 3 cm respectively.

OA = OB = 5 cm

O'A = O'B = 3 cm

OO' will be the perpendicular bisector of chord AB.

AC = CB

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.

Now, AC

AC = 3 m

^{2}= 25 - x^{2}= 25 - 4^{2}= 25 - 16 = 9AC = 3 m

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

### Solution 2

Let PQ and RS are two equal chords of a given circle and there are intersecting each other at point T.

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

OT = OT (common)

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)

VT = UT (by CPCT) ... (1)

It is given that

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

PV = RU ... ... ... ... (3)

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.

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

Let PQ and RS are two equal chords of a given circle and there are intersecting each other at point T.

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

OT = OT (common)

OVT OUT (RHS congruence rule)

OTV = OTU (by CPCT)

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)

Hence, the line joining the point of intersection to the centre makes equal angles with the chords.

### Solution 4

Let us draw a perpendicular OM on line AD.

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

Draw perpendiculars OA and OB on RS and SM respectively.

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

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

AR = AS = = 3cm

OR = OS = OM = 5 m (radii of circle)

In OAR

OA

OA

OA

OA = 4 m

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

RCS will be of 90

Area of ORS = OARS

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 = CMArea of ORS = OARS

RC = 4.8

RM = 2RC = 2(4.8)= 9.6

So, distance between Reshma and Mandip is 9.6 m.

### Solution 6

Given that AS = SD = DA

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

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

AB = OA + OB = (20 + 10) m = 30 m.

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. 9.3

### Solution 1

We may observe that

AOC = AOB + BOC

= 60

= 90

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.

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

In OAB

AB = OA = OB = radius

OAB is an equilateral triangle.

AB = OA = OB = radius

OAB is an equilateral triangle.

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

^{o}

AOB = 60

^{o}Now,

In cyclic quadrilateral ACBD

ACB + ADB = 180

ADB = 180

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

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.

PQR + PSR = 180

PSR = 180

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}(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}In POR

OP = OR (radii of same circle)

OP = OR (radii of same circle)

OPR = ORP (Angles opposite equal sides of a triangle)

OPR + ORP + POR = 180

^{o}(Angle sum property of a triangle)2 OPR + 160

2 OPR = 180

^{o}= 180^{o}2 OPR = 180

^{o}- 160^{o}= 20^{o}OPR = 10

^{o}### Solution 4

In ABC

BAC + ABC + ACB = 180

BAC + 69

^{o}(Angle sum property of a triangle)BAC + 69

^{o}+ 31^{o}= 180^{o}BAC = 180

BAC = 80

^{o}- 100ºBAC = 80

^{o}BDC = BAC = 80

^{o}(Angles in same segment of circle are equal)### Solution 5

In CDE

CDE + DCE = CEB (Exterior angle)

CDE + 20

CDE + DCE = CEB (Exterior angle)

CDE + 20

^{o}= 130^{o}CDE = 110

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

BAC = 110

^{o}But BAC = CDE (Angles in same segment of circle)

BAC = 110

^{o}### Solution 6

For chord CD

CBD = CAD (Angles in same segment)

CAD = 70

^{o}BAD = BAC + CAD = 30

BCD + BAD = 180

BCD + 100

BCD = 80

In ABC

^{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

AB = BC (given)

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

BCA = 30

We have BCD = 80

BCA + ACD = 80

30

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}ACD = 50

^{o}ECD = 50

^{o}### Solution 7

Let ABCD a cyclic quadrilateral having diagonals as BD and AC intersecting each other at point O.

(Consider BD as a chord)

BCD + BAD = 180

BCD = 180

^{o}(Cyclic quadrilateral)BCD = 180

^{o}- 90^{o}= 90^{o} (Considering AC as a chord)

ADC + ABC = 180

90

ABC = 90

Here, each interior angle of cyclic quadrilateral is of 90

^{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

Consider a trapezium ABCD with AB | |CD and BC = AD Draw AM CD and BN CD

In AMD and BNC

AD = BC (Given)

AMD = BNC (By construction each is 90

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

AMD BNC (RHS congruence rule)

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)

ADC = BCD (CPCT) ... (1)

BAD and ADC are on same side of transversal AD

BAD + ADC = 180

BAD + BCD = 180

This equation shows that the opposite angles are supplementary.

So, ABCD is a cyclic quadrilateral.

^{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

Join chords AP and DQ

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

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

Consider a ABC

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

ADC = 90

BDC = ADB + ADC = 90

Hence BDC is straight line and our assumption was wrong.

Thus, Point D lies on third side BC of ABC

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

In ABC

ABC + BCA + CAB = 180

90

BCA + CAB = 90

In ADC

^{o}(Angle sum property of a triangle)90

^{o}+ BCA + CAB = 180^{o}BCA + CAB = 90

^{o}... (1)In ADC

CDA + ACD + DAC = 180

90

ACD + DAC = 90

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

^{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

BCA + CAB + ACD + DAC = 180

^{o} (BCA + ACD) + (CAB + DAC) = 180

But it is given that

B + D = 90

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

So, it is a cyclic quadrilateral.

Consider chord CD.

Now, CAD = CBD (Angles in same segment)

^{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

Let ABCD be a cyclic parallelogram.

A + C = 180

We know that opposite angles of a parallelogram are equal

A = C and B = D

From equation (1)

A + C = 180

A + A = 180

2 A = 180

A = 90

Parallelogram ABCD is having its one of interior angles as 90

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.