NCERT Solutions for Class 9 Maths Chapter 10 Circles

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Written by Team Trustudies
Updated at 2021-02-15


NCERT solutions for class 9 Maths Chapter 10 Circles Exercise 10.1

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.

NCERT solutions for class 9 Maths Chapter 10 Circles Exercise 10.2

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: ?MON and ?PO?Q
image
Proof: In ?MON and ?PO?Q, we have,

MO = PO' ...(Radii of congruent circles)
NO = QO' ...(Radii of congruent circles)
and MN = PQ ...(Given)
By SSS criterion, we get,

?MON ? ?PO?Q
Hence, ?MON = ?PO?Q ...(By CPCT)

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., ?MON = ?PO?Q

image

To prove: MN = PQ

Proof:
In ?MON and ?PO?Q, we have,

MO = PO' ...(Radii of congruent circles)
NO = QO' ...(Radii of congruent circles)
and ?MON = ?PO?Q ...(Given)
By SAS criterion, we get,

?MON ? ?PO?Q

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

NCERT solutions for class 9 Maths Chapter 10 Circles Exercise 10.3

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)
imageimage
iii) No points in common (internally)iv) One point in common (externally)
imageimage
v) Two points in common
image

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.

Q2 ) Suppose you are given a circle. Give a construction to find its centre.



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.
image
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 OO’ intersect MN at P.

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

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

image

Proof:
In ?OMO? and ?ONO?, we have,

OM = ON ...(Radii of congruent circles)
O'M = O'N ...(Radii of congruent circles)
and OO' = O'O ...(Common)
By SSS criterion, we get,
?OMO? ? ?ONO?

Hence, ?MOO? = ?NOO? ...(By CPCT)
? ?MOP = ?NOP ...(i)(? ?MOO? = ?MOP, ?NOO? = ?NOP)

In ?MOP and ?NOP, we have,
OM = ON ...(Radii of congruent circles)
OM = OM ...(Common)
and ?MOP = ?NOP ...(from eq. (i))
By SAS criterion, we get,
?MOP ? ?NOP

Hence, MP = NP ...(By CPCT)
And ?MPO = ?NPO ...(ii)

But ?MPO + ?NPO = 180?
(Since, MPN is a straight line)
? 2 ?MPO = 90? ...(from (ii))
? ?MPO = 90?

We have,
MP = NP and ?MPO = ?NPO = 90?

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

NCERT solutions for class 9 Maths Chapter 10 Circles Exercise 10.4

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

image

In ?OO?A, we have,
AO?2+OO?2=32+42=9+16=25
? OA2=52=AO?2+OO?2

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

Now, Area of ?OO?A = 12 × O'A × OO'
= 12 × 3 × 4

Area of ?OO?A = 6 sq.cm ...(i)
Also, Area of ?OO?A = 12 × OO' × AM
= 12 × 4 × AM
Area of ?OO?A = 2 × AM ...(ii)

From eq. (i) and (ii), we get,
2 × 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 × AM = 2 × 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 OD?MN and OC?AB.Join OP

image

Proof:
DM = DN = (12 ) MN
(Since, Perpendicular from centre bisects the chord)
Similarly,
AC = CB = (12 ) AB
(Since, Perpendicular from centre bisects the chord)
? we get,
MD = BC and DN = AC ... (i)(Since, MN = AB)

In ?ODP and ?OPC,
OD = OC
(Equal chords of a circle are equidistant from the centre)
?ODP = ?OCP
and OP = OP ...(Common side)
? by RHS criterion of congruence,
?ODP ? ?OPC
Also, DP = PC ...(By CPCT) ...(ii)

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

MD + DP = BC + PC
? MP = PB
Also, On subtracting Equation (ii) and (i),we get,

DN + DP = AC + PC
? PN = AP

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.
image



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: ?OPC = ?OPB

Construction: Draw OC?RQ and OB?MN. Join OP.

Proof:
In ?OCP and ?OBP,
OC = OB
(Equal chords of a circle are equidistant from the centre)
?OCP = ?OBP
(Each angle is 90?)
and OP = OP ...(Common side)

Therefore, by RHS criterion of congruence,
?OCP ? ?OBP
Also, ?OPC = ?OPB ...(By CPCT)

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).
image



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.

image

Now, BC is the chord of the smaller circle and OP?BC.

Therefore, BC = PC ...(i)
Since, AD is a chord of the larger circle and OP?AD.

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.

image

Area of ?ORS = 12 × x × 5 = 5x2 ...(i)
(? in ?ORM, RM is a chord
? OP?RM)

In right angled triangle ?RON,
OR2=RN2+NO2
? 52=32+NO2
? NO2=25?9=16
?NO = 4cm
So, Area of ?ORS = 12 × RS × ON = 12 × 6 × 4 = 12 ...(ii)

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

? 5x2 = 12
? x = 245 .

Since, P is the mid- Point of RM,

Thus, RM = 2RP = 2 × 245
? RM = 485 = 9.6m

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

image

? DPQR is an equilateral traingle.

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 ?PQC,
PQ2=PC2+QC2
(By Pythagoras theorem)
? x2=PC2+(x2)2
?PC2=x2?(x2)2
? PC2=x2?x24=3x24
(Since, QC = 12 QR = x2 )
? PC = 3x2
Now, MC = PC - PM = 3x2 - 20
(Since, PM = radius)

Now, in ?QCM,
QM2=QC2+MC2
(By Pythagoras theorem)
? 202=(x2)2+(3x2?20)2
(Since, QM = radius)
? 400=x24+(3x2)2?203x+400
? 0=x2?203x
? x2=203x
? x=203

Hence, PQ = QR = PR = 203m.

NCERT solutions for class 9 Maths Chapter 10 Circles Exercise 10.5

Q1 ) In figure 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.
Let PQ = QR = PR = X
image



NCERT Solutions for Class 9 Maths Chapter 10 Circles


Answer :

Here, ?AOC = ?AOB + ?BOC = 60? + 30? = 90?

? Arc ABC makes 90? at the centre of the circle.

Thus, ?ADC = (12) ?AOC
(Since, The angle subtended by an arc at the centre is double the angle subtended by it any part of the circle)
? = (12 ) 90? = 45?

Hence, ?ADC = 45?.

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.

image

Given:
BC = OB = OC
?, ?OBC is an equilateral traingle.
? ?BOC = 60?
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.

? ?BAC = (12 ) ?BOC
? ?BAC = (12 ) 60? = 30?

Also, here, ABMC is a cyclic quadrilateral.

? ?BAC + ?BMC = 180?
(Since, in a cyclic quadrilateral the sum of opposite angles is 180?).
? ?BMC = 180? - 30? = 150?

Hence, the angle subtended by the chord at a point on the minor arc and also at a point on the major arc is 150?.

Q3 ) In figure, ?PQR = 100? ,where P, Q and R are points on a circle with centre O. Find ?OPR.
image



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,

?PQR = 2 ?PQR = 200?
(given: ?PQR = 100?)

? in ?OPR,
?POR = 360? - 200? = 160? ...(i)

Again, in ?OPR,
OP = OR ...(Radii of the circle)
?OPR = ?ORP
(By property of isosceles traingle)

In ?POR,
?OPR + ?ORP + ?POR = 180? ...(ii)
(Sum of internal angles of triangle)

? from eq.(i) and (ii), we get,

?OPR + ?OPR + 160? = 180?
? 2 ?OPR = 180? - 160? = 20?
Therefore, ?OPR = 10?.

Q4 ) In figure, ?ABC = 69?, ?ACB = 31?. Find ?BDC.
image



NCERT Solutions for Class 9 Maths Chapter 10 Circles


Answer :

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

?BDC = ?BAC ...(i)

Now, in ?ABC,

?BAC + ?ABC + ?ACB = 180?
? ?BAC + 69? + 31? = 180?
??ABC + 100? = 180?
? ?ABC = 180? - 100?
? ?BAC = 80?
? ?BDC = 80? ...(from (i))

Q5 ) In figure, A, B and C are four points on a circle. AC and BD intersect at a point E such that ?BEC = 130? and ?ECD = 20?. Find ?BAC.
image



NCERT Solutions for Class 9 Maths Chapter 10 Circles


Answer :

In ?ABC,

?AED = 180? -130? = 50?
(linear pair of angles)

? ?CED = ?AED = 50?
(Vertically opposite angles)

Also, ?ABD = ?ACD
(Since, the angles in the same segment are equal)
? ?ABE = ?ECD
? ?ABE = 180? ...(ii)

Now, in ?CDE,
?BAC + 20? + 50? = 180?
(from eq.(i) and (ii))

? ?BAC + 70? = 180?
? ?BAC = 180? - 70?
? ?BAC = 110?

Q6 ) ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If ?DBC = 70?, ?BAC = 30?, Find ?BCD. Further, if AB = BC, find ?EDC.



NCERT Solutions for Class 9 Maths Chapter 10 Circles


Answer :

image

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

?BDC = ?BAC
? ?BDC = 30?.

In ?DBC, we have,

?BDC + ?DBC + ?BCD = 180?
? 30? + 70? + ?BCD = 180?
? 100? + ?BCD = 180?
? ?BCD = 180? - 100?
? ?BCD = 80?

If AB= BC, ?BCA = ?BAC = 30?
(Since, Angles opposite to equal sides in a traingle are equal)

Now,
?ECD = ?BCD - ?BCA = 80? - 30?
(Given ?BCD = 80? and ?BCA = 30?)

? ?ECD = 30?
? ?EDC = 30?

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.

image

To prove: Quadrilateral NQPM is a rectangle.

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

Now, ON = OP = (12 ) NP
and OM = OQ = (12 ) MQ
? NP = MQ

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

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

Q8 ) If the non-parallel sides of a trapezium are equal, prove that it is cyclic.



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 SM?PQ and RN?PQ.

image

To prove: ABCD is cyclic trapezium.

Proof:
In ?SMP and ?RNQ, we have,

SP = RQ ...(Given)
SM = RN
(Distance between two parallel twolines is always equal)

and ?SMP = ?RNQ
(Each 90?)
By RHS criterion, we get,

?SMP ? ?RNQ

Hence, ?P = ?Q ...(By CPCT)

Also, ?PSM =