Q1 )
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.

NCERT Solutions for Class 9 Maths Chapter 10 Circles

Answer :

i) The centre of a circle lies in __interior__ of the circle.

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

iii) The longest chord of a circle is a __diameter__ of 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 and __chord__ of the circle.

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

Q2 )
Write True or False. Give reason 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.

NCERT Solutions for Class 9 Maths Chapter 10 Circles

Answer :

i) True, Because all points are equidistant from the centre to the circle.

ii) False, Because circle has infinitely may equal chords can be drawn.

iii) False, Because all three arcs are equal, so there no difference between the major and
minor arcs.

iv) True, By the definition of diameter, that diameter is twice the radius.

v) False, Because the sector is the region between two radii and an arc.

vi) True, Because circle is a part of the plane figure.

Q1 ) Recall that two circles are congruent, if they have the same radii. Prove that equal chords of congruent circles subtend equal angles at their centers.

NCERT Solutions for Class 9 Maths Chapter 10 Circles

Answer :

Given: MN and PO are two equal chords of two congruent circles with centre at O and
O'.

To prove:

Proof: In

MO = PO' ...(Radii of congruent circles)

NO = QO' ...(Radii of congruent circles)

and MN = PQ ...(Given)

By SSS criterion, we get,

Hence,

Q2 ) Prove that, if chords of congruent circles subtend equal angles at their centres, then the chords are equal.

NCERT Solutions for Class 9 Maths Chapter 10 Circles

Answer :

Given:

MW and PQ are two chords of congruent circles such that angles subtended by
these chords at the centres O and O' of the circles are equal i.e.,

To prove: MN = PQ

Proof:

In

MO = PO' ...(Radii of congruent circles)

NO = QO' ...(Radii of congruent circles)

and

By SAS criterion, we get,

Hence, MN = PQ ...(By CPCT)

Q1 )
Draw different pairs of circles. How many points does each pair have in common?

what is the maximum number of common points?

NCERT Solutions for Class 9 Maths Chapter 10 Circles

Answer :

Different pairs of circles are as follows:

i) No points in common (externally) | (ii) One point in common (internally) |
---|---|

iii) No points in common (internally) | iv) One point in common (externally) |

v) Two points in common | |

From figures, it is obvious that these pairs many have 0 or 1 or 2 points in common.

Hence, a pair of circles cannot intersect each other at more than two points.

NCERT Solutions for Class 9 Maths Chapter 10 Circles

Answer :

Steps of construction:

1. Taking three points P, Q and R on the circle.

2. Join PQ and QR.

3. Draw MQ and NS, respectively the perpendicular bisectors of PQ and RQ, which intersect each other at O.

Hence, O is the centre of the circle.

Q3 ) If two circles intersect at two points, prove that their centres lie on the perpendicular bisector of the common chord.

NCERT Solutions for Class 9 Maths Chapter 10 Circles

Answer :

Given:

Two circles with centres O and O' intersect at two points M and N so that MN is the common chord of the two circles and OO' is the line segment joining the centres of the two circles. Let OOâ€™ intersect MN at P.

To prove: OO' is the perpendicular bisector of MN.

Construction:

Draw line segments OM, ON, O'M and O'N.

Proof:

In

OM = ON ...(Radii of congruent circles)

O'M = O'N ...(Radii of congruent circles)

and OO' = O'O ...(Common)

By SSS criterion, we get,

Hence,

In

OM = ON ...(Radii of congruent circles)

OM = OM ...(Common)

and

By SAS criterion, we get,

Hence, MP = NP ...(By CPCT)

And

But

(Since, MPN is a straight line)

We have,

MP = NP and

Hence, it is proved that OO' is the perpendicular bisector of MN.

Q1 ) 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.

NCERT Solutions for Class 9 Maths Chapter 10 Circles

Answer :

Let O and O' be the centres of the circles of radii 5 cm and 3 cm, respectively.

Let AB be their common chord.

Given: OA = 5cm, O'A = 3cm and OO' = 4cm

In

Therefore, OO'A is a right angled traingle and right angled at O'

Now, Area of

=

Area of

Also, Area of

=

Area of

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

2 Ã— AM = 6

Therefore, AM = 3 cm.

Since, when two circles intersect at two points, then their centres lie on the perpendicular bisector of the common chord.

Thus, AB= 2 Ã— AM = 2 Ã— 3 = 6cm.

Therefore, the length of the common chord is 6cm.

Q2 ) 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.

NCERT Solutions for Class 9 Maths Chapter 10 Circles

Answer :

Given:

MN and AB are two chords of a circle with centre O, AB and MN intersect at P
and MN = AB.

To prove: MP = PB and PN = AP

Construction: Draw

Proof:

DM = DN = (

(Since, Perpendicular from centre bisects the chord)

Similarly,

AC = CB = (

(Since, Perpendicular from centre bisects the chord)

MD = BC and DN = AC ... (i)(Since, MN = AB)

In

OD = OC

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

and OP = OP ...(Common side)

Also, DP = PC ...(By CPCT) ...(ii)

On adding Equations (i) and (ii), we get,

MD + DP = BC + PC

Also, On subtracting Equation (ii) and (i),we get,

DN + DP = AC + PC

Hence, MP = PB and PN = AP are proved.

Q3 )
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.

NCERT Solutions for Class 9 Maths Chapter 10 Circles

Answer :

Given:

RQ and MN are chords of a circle with centre O. MN and RQ intersect at P and
MN = RQ.

To prove:

Construction: Draw

Proof:

In

OC = OB

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

(Each angle is

and OP = OP ...(Common side)

Therefore, by RHS criterion of congruence,

Also,

Q4 )
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 figure).

NCERT Solutions for Class 9 Maths Chapter 10 Circles

Answer :

Let OP be the perpendicular from O on linel . Since, the perpendicular from the centre
of a circle to a chord bisects the chords.

Now, BC is the chord of the smaller circle and

Therefore, BC = PC ...(i)

Since, AD is a chord of the larger circle and

Therefore, AP = PD ...(ii)

On subtracting Eq. (i) from Eq.(ii), we get,

AP - BP = PD - PC

Thus, AB = CD

Hence, it is proved.

Q5 )
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, and Mandip to Reshma.

If the distance between Reshma and Salma and between Salma and Mandip is 6 m each, what is the distance between Reshma and Mandip?

NCERT Solutions for Class 9 Maths Chapter 10 Circles

Answer :

Let O be the centre of the circle and Reshma, Salma and Mandip are represented by
the points R, S and M, respectively. Let RP = x m.

Area of

(

In right angled triangle

So, Area of

From Equations (i) and (ii), we get,

Since, P is the mid- Point of RM,

Thus, RM = 2RP = 2 Ã—

Hence, the distance between Reshma and Mandip is 9.6m.

Q6 )
A circular park of radius 20 m 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.

NCERT Solutions for Class 9 Maths Chapter 10 Circles

Answer :

Let Ankur, Syed and David standing on the point P, Q and R.

Let PQ = QR = PR = X

Drawn altitudes PC, QD and RN from vertices to the sides of a traingle and intersect these altitudes at the centre of a circle M.

As PQR is an equilateral, therefore these altitudes bisects their sides.

In

(By Pythagoras theorem)

(Since, QC =

Now, MC = PC - PM =

(Since, PM = radius)

Now, in

(By Pythagoras theorem)

(Since, QM = radius)

Hence, PQ = QR = PR =

Q1 )
In figure A, B and C are three points on a circle with centre O such that

Let PQ = QR = PR = X

NCERT Solutions for Class 9 Maths Chapter 10 Circles

Answer :

Here,

Thus,

(Since, The angle subtended by an arc at the centre is double the angle subtended by it any part of the circle)

Hence,

Q2 ) 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.

NCERT Solutions for Class 9 Maths Chapter 10 Circles

Answer :

Let BC be chord, which is equal to the radius. Join OB and OC.

Given:

BC = OB = OC

But, we know that,

The angle subtended by an arc at the centre is double the angle subtended by it at any part of the circle.

Also, here, ABMC is a cyclic quadrilateral.

(Since, in a cyclic quadrilateral the sum of opposite angles is

Hence, the angle subtended by
the chord at a point on the minor arc and also at a point on the major arc is

Q3 )
In figure,

NCERT Solutions for Class 9 Maths Chapter 10 Circles

Answer :

Since, the angle subtended by the centre is double the angle subtended by
circumference, we get,

(given:

Again, in

OP = OR ...(Radii of the circle)

(By property of isosceles traingle)

In

(Sum of internal angles of triangle)

Therefore,

NCERT Solutions for Class 9 Maths Chapter 10 Circles

Answer :

As we know that, the angles in the same segment are equal.

We get,

Now, in

Q5 )
In figure, A, B and C are four points on a circle. AC and BD intersect at a point E
such that

NCERT Solutions for Class 9 Maths Chapter 10 Circles

Answer :

In

(linear pair of angles)

(Vertically opposite angles)

Also,

(Since, the angles in the same segment are equal)

Now, in

(from eq.(i) and (ii))

Q6 )
ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If

NCERT Solutions for Class 9 Maths Chapter 10 Circles

Answer :

As, Angles in the same segment are equal, we get,

In

If AB= BC,

(Since, Angles opposite to equal sides in a traingle are equal)

Now,

(Given

Q7 ) If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.

NCERT Solutions for Class 9 Maths Chapter 10 Circles

Answer :

Given:

Diagonals NP and QM of a cyclic quadrilateral are diameters of the circle
through the vertices M, P, Q and N of the quadrilateral NQPM.

To prove: Quadrilateral NQPM is a rectangle.

Proof:

ON = OP = OQ = OM ...(Radii of circle)

Now, ON = OP = (

and OM = OQ = (

Hence, the diagonals MPQN are equal and bisect each other.

Hence, it is proved that quadrilateral NQPM is a rectangle.

NCERT Solutions for Class 9 Maths Chapter 10 Circles

Answer :

Given:

Non-parallel sides PS and QR of a trapezium PQRS are equal.

Construction:

Draw

To prove: ABCD is cyclic trapezium.

Proof:

In

SP = RQ ...(Given)

SM = RN

(Distance between two parallel twolines is always equal)

and

(Each

By RHS criterion, we get,

Hence,