NCERT Exemplar Class 10 Maths solutions Chapter 9 Circles

NCERT exemplar Class 10 Maths solutions chapter 9 extends the learning of circles. In earlier classes, we have learned about circles; here, we will learn about tangents at any point of the circle. This chapter covers the definition of The NCERT exemplar Class 10 Maths chapter 9 solutions have been developed by our experienced Mathematics division subject matter experts to give students a quick and easy understanding of NCERT Class 10 Maths.

These Class 10 Maths NCERT exemplar chapter 9 solutions on the chapter circles are exceptionally well structured and detailed. The CBSE Syllabus for Class 10 is the building block for these NCERT exemplar Class 10 Maths solutions chapter 9.

Question:1

If radii of two concentric circles are 4 cm and 5 cm, then the length of each chord of one circle which is tangent to the other circle is
(A) 3 cm (B) 6 cm (C) 9 cm (D) 1 cm

Answer: (B) 6 cm

Solution
According to question

Here A is the center and AB = 4 cm radius of small circle and AD = 5 cm radius of large circle.
We have to find the length of CD
ABD is a right angle triangle.
Hence use Pythagoras theorem in ABD










Hence length of chord is 6 cm.

Question:2

In Figure, if ∠AOB = 125°, then ∠COD is equal to

(A) 62.5° (B) 45° (C) 35° (D) 55°

Answer: (D)
Solution

Given : ∠AOB = 125°
Let ∠COD = x°
As we know that the sum of opposite sides angles of a quadrilateral circumscribing a circle is equal to 180°.
Hence
∠AOB + ∠COD = 180°
125° + ∠COD = 180°
x° = 180° – 125° ( ∠COD = x°)
x° = 55°
Hence, ∠COD = 55°

Question:3

In Figure, AB is a chord of the circle and AOC is its diameter such that ACB = 50°. If AT is the tangent to the circle at the point A, then BAT is equal to

(A) 65° (B) 60° (C) 50° (D) 40°

Answer: (C)
Solution

Given ACB = 50°.
We know that the angle subtended by a diameter is right.
Hence B = 90°
We know that sum of interior angle of a triangle is 180°.
In ABC




That is

Question:4

From a point P which is at a distance of 13 cm from the centre O of a circle of radius 5 cm, the pair of tangents PQ and PR to the circle are drawn. Then the area of the quadrilateral PQOR is
(A) 60 cm² (B) 65 cm² (C) 30 cm² (D) 32.5 cm²

Answer:

(A) 60 cm²
Solution
According to question

Given OP = 13 cm
OQ = OR = radius = 5 cm
In POQ, ( PQ is tangent)
Using Pythagoras theorem in POQ
( because H² = B² + P²)




Area of POQ = × perpendicular × base


Area of quadrilateral = 2 × area of POQ
= 2 × 30 = 60 cm²

Question:5

At one end A of a diameter AB of a circle of radius 5 cm, tangent XAY is drawn to the circle. The length of the chord CD parallel to XY and at a distance 8 cm from A is
(A) 4 cm (B) 5 cm (C) 6 cm (D) 8 cm

Answer:

(D) 8 cm
Solution
According to questions

Given AO = OB = 5 cm
Distance between XY and CD = 8 cm
Since D is the center of the circle and CD is chord. If we join OD it become the radius of circle that is OD = 5cm
AZ = 8 cm (given)
AZ = AO + OZ
8 = 5 + OZ
OZ = 3cm
In ODZ, use Pythagoras theorem







Length of chord CD = 8 cm

Question:6

In Figure, AT is a tangent to the circle with centre O such that OT = 4 cm and ∠OTA = 30°. Then AT is equal to

(A) 4 cm (B) 2 cm (C) (D)

Answer: (C)

Solution

Given ∠OTA = 30° , OT = 4cm
Join OA, since AT is tangent.
Hence it is perpendicular to OA
In OAT

Question:7

In Figure, if O is the centre of a circle, PQ is a chord and the tangent PR at P makes an angle of 50° with PQ, then ∠POQ is equal to


(A) 100° (B) 80° (C) 90° (D) 75°

Answer: (A)

Solution
Given :
We know that tangent is perpendicular to radius.
Hence
OPR =
OPR = OPQ + QPR
= OPQ +
OPQ =
OPQ = OQD
Hence , OPQ =
We know that the sum of interior angles of a triangle is
In
O+Q +P =
O +40+40 = 180
o =
Hence OPQ =

Question:8

In Figure, if PA and PB are tangents to the circle with centre O such that APB = 50°, then OAB is equal to

(A) 25° (B) 30° (C) 40° (D) 50°

Answer: (A) 25°

Solution
Given : APB = 50°
We know that length of tangents drawn from an external point is equal
Hence, PA = PB
Since, PA = PB
Let PAB = PBA = x⁰
In PAB
p+A+B = ( Sum of interior angles of a tangent is )



PAB = PBA =
PAO = ( tangent is perpendicular to radius)
PAO = PAB +OAB

Question:9

If two tangents inclined at an angle 60° are drawn to a circle of radius 3 cm,then length of each tangent is equal to
(A) (B) 6 cm (C) 3 cm (D)

Answer:

(D)
Solution
According to question

Given OQ = OR = 3 cm (Radius)

Draw line OP which bisect . That is

In




Here PQ = PR
Hence , PQ = PR=

Question:10

In Figure, if PQR is the tangent to a circle at Q whose centre is O, AB is a chord parallel to PR and BQR = 70°,then AQB is equal to

( A) 20° (B) 40° (C) 35° (D) 45°

Answer:

Answer(B)
Solution
Given :,
Since DQ perpendicular to PR




Since DQ bisect
Hence ,

Question:1

Write ‘True’ or ‘False’ and justify your answer in each of the following :
If a chord AB subtends an angle of at the centre of a circle, then angle between the tangents at A and B is also .

Answer:False

Solution

Here CA and CB are the two tangents which is drawn on chord AB and also we know that tangent and radius are perpendicular to each other.
i.e.,
In quadrilateral, ABCD

[Sum of interior angles of a quadrilateral is ]



Here we conclude that angle between the tangents at A and B is .
Therefore the given statement is False.

Question:2

Write ‘True’ or ‘False’ and justify your answer in each of the following :The length of tangent from an external point on a circle is always greater than the radius of the circle.

Answer:

False
Solution
Case-I – When external point P is very close to circle

Here C is the center of circle. Let this radius is 5 and distance of Px and Py is 3.Or we can say that Case-I contradict the given statement.
Case-II – When external point P is far the circle.

Here radius of the given circle is 5 cm and let the length of Px and Py is 10 cm.
Hence according to Case-II and given statement is True.
Hence from the above two cases we conclude that the tangent’s length is depend on the distance of external point from the circle.
Therefore given statement is False because length of tangent from an external point of a circle may or may not be greater than the radius of the circle.
Therefore the given statement is False.

Question:3

Write ‘True’ or ‘False’ and justify your answer in each of the following :The length of tangent from an external point P on a circle with centre O is always less than OP.

Answer:

True
Solution


Here O is the center of given circle and PA is tangent which is drawn from an external point P. OA is radius of circle.We know that tangent and radius is always perpendicular to each other.

is a right angle triangle Use Pythagoras theorem in
…..(i)
From equation (i) we can say that PA is always less than PO.
In other words we can say that the length of hypotenuse is always greater than the length of perpendicular in a right angle triangle.
i.e., OP PA
Hence the given statement is True.

Question:4

Write ‘True’ or ‘False’ and justify your answer in each of the following :The angle between two tangents to a circle may be .

Answer:

True
Solution
It is possible that the angles of two line may be in only two conditions.
1. When lines are parallel
2.When both the lines are coincide.
Hence the given statement is True
This may be possible only when tangents are parallel or when both the tangents are coincide.

Question:5

Write ‘True’ or ‘False’ and justify your answer in each of the following :If angle between two tangents drawn from a point P to a circle of radius a and centre O is , then

Answer:

True
Solution

Given
Draw line OP from point O to P which bisect .
i.e ,
In





Hence the given statement is True.

Question:6

Write ‘True’ or ‘False’ and justify your answer in each of the following :If angle between two tangents drawn from a point P to a circle of radius a and centre O is , then

Answer:

False
Solution

Given
Draw line OP from point O to P which bisect P. Which bisect
i.e,
In




OP = 2a
Hence the value of OP = 2a
Hence the given statement is False.

Question:7

Write ‘True’ or ‘False’ and justify your answer in each of the following :The tangent to the circumcircle of an isosceles triangle ABC at A, in which AB = AC, is parallel to BC.

Answer:

Answer: true
Solution

Given ABC is a isosceles triangle and AB = AC.
To Prove :
Proof : AB = AC (Given)

[ angle between chord of a circle and tangent is equal to angle made by chord in alternate segment]
Also


i.e ,
Hence true

Question:8

Write ‘True’ or ‘False’ and justify your answer in each of the following
If a number of circles touch a given line segment PQ at a point A, then their centres lie on the perpendicular bisector of PQ.

Answer:

False
Solution
According to question.

Here C₁, C₂, C₃ are the circle with center O₁, O₂, O₃ respectively.
C₁, C₂, C₃ touches line PQ at point A. Here PQ is the tangent at each circle.
If we join O₁, O₂, O₃ to point A then the line is perpendicular to line PQ because if we draw a line from the center of circle at any point then the tangent at that point is perpendicular to the radius.
But here line joining the centers is not bisecting the line PQ because it depend on the length of PQ.
Hence the given statement is False.

Question:9

Write ‘True’ or ‘False’ and justify your answer in each of the following
If a number of circles pass through the end points P and Q of a line segment
PQ, then their centres lie on the perpendicular bisector of PQ.

Answer:

True
Solution
According to question.

Here C₁, C₂ circles pass through the point P and Q.
We know that the perpendicular bisector of chord of a circle is always passes through the center of the circle. Hence the perpendicular bisector of line PQ passes through the center of circles of C₁, C₂.
Hence the given statement is True.

Question:10

Write ‘True’ or ‘False’ and justify your answer in each of the following
AB is a diameter of a circle and AC is its chord such that BAC = 30°. If the tangent at C intersects AB extended at D, then BC = BD.

Answer:

True
Solution: First of all we solve the question according to give conditions. If we able to prove it then it will be true otherwise it will be false.
Given :BAC =
Diagram : Construct figure according to given conditions then join BC and OC.

To Prove : BC = BD
Proof :BAC = (Given)

[ angle between chord and tangent id equal to the angle made by chord in alternate segment]

[ Radius and tangent’s angle is always ]
In OAC
OA = OC (both are radius of circle)

[opposite angles of an isosceles triangle is equal]


In

[ sum of interior angle of a trianglE ]



In BCD we conclude that
and
[sides which is opposite to equal angles is always equal]
Hence Proved.
Hence the given statement is true.

Question:1

Out of the two concentric circles, the radius of the outer circle is 5 cm and the chord AC of length 8 cm is a tangent to the inner circle. Find the radius of the inner circle.

Answer:

Radius = 3 cm
Solution
According to question

Given : AC = 8 CM
OC = 5 CM
AC = AB+BC
8 = BC+BC ( AB =BC)
BC = 4CM
In using Pythagoras theorem





Hence radius of inner circle is 3 cm.

Question:2

Two tangents PQ and PR are drawn from an external point to a circle with centre O. Prove that QORP is a cyclic quadrilateral.

Answer:

Solution
According to question

To Prove : QORP is a cyclic quadrilateral.
OQ PQ, OR PR ( PQ, PR are tangents)
Hence,
We know that sum of interior angles of quadrilateral is
[Given (i)]


Here we found that sum of opposite angles of quadrilateral is
Hence QORP is a cyclic quadrilateral.
Hence proved

Question:3

If from an external point B of a circle with centre O, two tangents BC andBD are drawn such that DBC = , prove that BC + BD = BO, i.e., BO = 2BC.

Answer:

Solution
According to question

Given :
To Prove : BC + BD = BO, i.e., BO = 2BC
Here OC BC, OD BD ( BC, BD are tangents)
Join OB which bisect
In







Hence Proved

Question:4

Prove that the centre of a circle touching two intersecting lines lies on the angle bisector of the lines.

Answer:

Solution

Here PQ and PR are tangents and O is the center of circle.
Let us join OQ and OR.
Here
( tangent from exterior point is perpendicular to the radius through the point of contact)
In PQO andPRO
(Radius of circle)
(Common side)
Hence, [RHS interior]
Hence, [By CPCT]
Hence O lie on angle bisector of
Hence Proved.

Question:5

In Figure, AB and CD are common tangents to two circles of unequal radii. Prove that AB = CD.

Answer:

Solution
To Prove : AB = CD
Extend AB and CD then they meet at point E.

Here EA = EC ( length of tangent drawn from same point is equal)
Also EB = ED ( length of tangent drawn from some point is equal)
AB = AE - BE
CD = CE -DF
From equal axiom if equal are subtracted from equal then the result is equal.

Hence Proved

Question:6

In Figure, AB and CD are common tangents to two circles of unequal radii.

In above question, if radii of the two circles are equal, prove that AB = CD.

Answer:

Solution
To Prove AB = CD
According to question

It is given that radius of both circles are equal
Hence, OA = OC = PB = PD
Here,
( tangent at any point is perpendicular to the radius at the point of contact)
Hence ABCD is a rectangle.
Opposite side of a rectangle are equal

Hence Proved.

Question:7

In Figure, common tangents AB and CD to two circles intersect at E. Prove that AB = CD.

Answer:

Solution
To Prove : AB = CD
We know that the length of tangents drawn from some point to a circle is equal.
So, EB = ED
AE = CE
Here AB = AE + EB
CD = CE + ED
From Euclid axiom is equals are added in equals then the result is also equal.
AB = CD
Hence Proved

Question:8

A chord PQ of a circle is parallel to the tangent drawn at a point R of the circle. Prove that R bisects the arc PRQ.

Answer:

Solution
According to question

To Prove : R bisects the arc PRQ
Here …..(i) ( Alternate interior angles)
We know that angle between tangent and chord is equal to angle made by chord in alternate segment.
…..(ii)
From equation (i) and (ii)

We know that sides opposite to equal angles and equal.
PR = QR
Hence R bisects the arc PRQ
Hence Proved

Question:9

Prove that the tangents drawn at the ends of a chord of a circle make equal angles with the chord.

Answer:

Solution
According to question

To Prove :
X₁, Y₁, X₂, Y₂ are tangents at point A and B respectively.
Take a point C and join AC and AB.
Now …..(i) (Angle in alternate segment)
Similarly,
…..(ii)
From equation (i) and (ii)
…..(iii)
From equation (3)

Hence Proved

Question:10

Prove that a diameter AB of a circle bisects all those chords which are parallel to the tangent at the point A.

Answer:

Solution
According to question

Let us take a chord EF || XY
Here
( tangent at any point of the circle is perpendicular to the radius through the point of contact)
(Corresponding angles)

Thus AB bisects EF
Hence AB bisects all the chords which are parallel to the tangent at the point A.
Hence Proved

Question:1

If a hexagon ABCDEF circumscribe a circle, prove that AB + CD + EF = BC + DE + FA.

Answer:

Solution
Given ABCDEF hexagon circumscribe a circle.

To Prove : AB + CD + EF = BC + DE + FA
Proof : Here AR = AS [ length of tangents drawn from a point are always equal]
Similarly,
BS = BT
CT = CU
DU = DV
EV = EW
FW = FR
Now AB + CD + EF = (AS + SB) + (CU + UD) + (EW + WF)
= (AR + BT) + (CT + DV) + (EV + FR)
= (AR + FR) + (BT + CT) + (DV + EV)
Now according to Euclid’s axiom when equals are added in equals then the result is also equal
AB + CD + EF = AF + BC + DE
i.e., AB + CD + EF = BC + DE + FA
Hence Proved

Question:2

Let s denote the semi-perimeter of a triangle ABC in which BC = a, CA = b, AB = c. If a circle touches the sides BC, CA, AB at D, E, F, respectively, prove that BD = s – b.

Answer:

Solution
Given : BC = a, CA = b, AB = c

Here, AF = AE = Z1 ( tangents drawn from an external point to the circle are equal in length)


Here AB + BC + CA = c + a + b
(AB + FB) + (BC + DC) + (CE + EA) = a + b + c








-
Hence Proved

Question:3

From an external point P, two tangents, PA and PB are drawn to a circle with centre O. At one point E on the circle tangent is drawn which intersects PA and PB at C and D, respectively. If PA = 10 cm, find the perimeter of the triangle PCD.

Answer:

In this figure
CE = CA [ Tangents from an external point to a circle are equal in length]
Similarly,
DE = DB and PB = PA
Perimeter of DPCD
= PC + CD + PD
= PC + CE + ED + PD ( CD = CE + ED)
= PC + CA + DB + PD
= PA + PB [ PC + CA = PA and DB + PD = PB]
= PA + PA [ PB = PA]
= 2PA
= 2 × 10 [ PA = 10]
= 20 cm

Question:4

If AB is a chord of a circle with centre O, AOC is a diameter and AT is the tangent at A as shown in Figure. Prove that BAT = ACB

Answer:

Solution
Here [ AC is a diameter line, Angle in semi circle formed, is ]
In ABC

[ sum of interior angles of a triangle is ]
......(I)
We know that diameter of a circle is perpendicular to the tangent.



Equate equation (i) and (ii) we get


Hence Proved

Question:5

Two circles with centres O and O' of radii 3 cm and 4 cm, respectively intersect at two points P and Q such that OP and O'P are tangents to the two circles. Find the length of the common chord PQ.

Answer:

4.8 cm
Solution

Given : Radii of two circles are OP = 3 cm and and intersection point of two circles are P and Q. Here two tangents drawn at point P are OP and P


Also apply Pythagoras theorem in we get




Equate equation (i) and (ii) we get









Put x = 1.8 in equation (i) we get

Question:6

In a right triangle ABC in which B = , a circle is drawn with AB as diameter intersecting the hypotenuse AC and P. Prove that the tangent to the circle at P bisects BC.

Answer:

Solution

Let O be the center of the given circle.
Suppose P meet BC at point R
Construction: Join point P and B
To Prove: BR = RC
Proof: [Given]
In ABC
[ Sum of interior angle of a triangle is 180°]



Also [ tangent and chord made equal angles in alternate segment]

[ angle in semi circle formed is 90°]




Equal equation (i) and (ii) we get


[Side opposite to equal angles are equal]
Also, PR = BR …..(iv) [ tangents drawn to a circle from external point are equal]
From equation (iii) and (iv)
BR = RC
Hence Proved

Question:8

AB is a diameter and AC is a chord of a circle with centre O such that BAC = . The tangent at C intersects extended AB at a point D. Prove that BC = BD.

Answer:

Solution

Given AB is a diameter and AC is a chord of circle with center O.

To Prove : BC = BD
Construction : Join B and C
Proof :[Angle is alternate segment]


[Angle in semi circle formed is 90°]
In ABC
[Sum of interior angles of a triangle is 180°]



Also , [Linear pair]



In


From equation (i) and (ii)

[ Sides opposite to equal angles are equal]
Hence Proved

Question:9

Prove that the tangent drawn at the mid-point of an arc of a circle is parallelto the chord joining the end points of the arc.

Answer:

Solution

Let mid-point of arc is C and DCE be the tangent to the circle.
Construction: Join AB, AC and BC.
Proof: In ABC
AC = BC

[ sides opposite to equal angles are equal]
Here DCF is a tangent line
[ angle in alternate segments are equal]
…..(ii) [From equation (i)]
But Here and are alternate angels.
equation (ii) holds only when AB||DCE.
Hence the tangent drawn at the mid-point of an arc of a circle is parallel to the chord joining the end points of the arc.
Hence Proved.

Question:10

In Figure, the common tangent, AB and CD to two circles with centres O and O' intersect at E. Prove that the points O, E, O' are collinear.

Answer:

Solution
Construction : Join AO and OS
and

In and
[Radius are equal]
[Common side]
ED = EB [Tangent drawn from an external point to the line circle are equal to length]
[By SSS congruence criterion]

i.e., is bisector of
Similarly OE is bisector of
In quadrilateral


[ is cyclic quadrilateral]
[ AB is a straight line]
[From equation (ii)]


Similarly
From equation (ii)
Dividing both side by 2



Similarly
Dividing both side by 2



Now


[from equation (iii) and (iv)]




So,OEO’ is straight line
O, E and are collinear.
Hence Proved

Question:11

In Figure. O is the centre of a circle of radius 5 cm, T is a point such that OT = 13 cm and OT intersects the circle at E. If AB is the tangent to the circle at E, find the length of AB.

Answer:

Answer
Given: Radius = 5 cm, OT = 13 cm
[ PT is tangent]
Using Pythagoras theorem OPT





PT and QT are tangents from same point
PT = QT = 12 cm
AT = PT – PA
AT = 12 – PA ..…(i)
Similarly BT = 12 – QB ..…(ii)
Since PA, PF and BF, BQ are tangents from point A and B respectively.
Hence, PA = AE ..…(iii)
BQ = BE ..…(iv)
AB is tangent at point E
Hence OE AB



ET = OT – OF
ET = 13 – 5
ET = 8 cm
In AET, using Pythagoras theorem


[using (i) and (ii)]






Similiraly

Question:12

The tangent at a point C of a circle and a diameter AB when extended intersect at P. If PCA = , find CBA [see Figure].

Answer:
Given :
Here [ PC is tangent]

Here OC = OA (Radius)
[ Sides opposite to equal angles are equal]
We know that PC is tangent and angles in alternate segment are equal.
Hence
In
[Interior angles sum of triangle is 180°]
[using (i)]
…..(ii)
Here
( using (ii))
…..(iii)
In
[using (iii)]


Hence

Question:13

If an isosceles triangle ABC, in which AB = AC = 6 cm, is inscribed in a circle of radius 9 cm, find the area of the triangle.

Answer:

Answer
Solution
According to question

In and
AB = AC [Given]
BO = CO [Radius]
AO = AO [Common side]
[By SSS congruence Criterion]
[CPCT]
In and
AB = AC [given]

AD = AD [common side]
[By SAS congruence Criterion]
…..(i) [CPCT]
…..(ii)
From (i) and (ii)

OA is a perpendicular which bisects chord BC
Let AD = x, then OD = 9 – x
Use Pythagoras in ADC


…..(iii)
In ODC using Pythagoras theorem

…..(iv)
From (iii) and (iv)






i.e., AD = 2 cm, OD = 9 – 2 = 7 cm
Put value of x in (iii)






Areao of

Question:14

A is a point at a distance 13 cm from the centre O of a circle of radius 5 cm.AP and AQ are the tangents to the circle at P and Q. If a tangent BC is drawn at a point R lying on the minor arc PQ to intersect AP at B and AQ at C, find the perimeter of the ABC.

Answer:

Answer 24 cm
Solution
Let us make figure according to question


Given : OA = 13 cm, Radius = 5 cm
Here AP = AQ (tangent from same point)
OP PA, OQ QA ( AP, AQ are tangents)
In OPA using Pythagoras theorem




........(i)
Perimeter of ABC = AB + BC + CA
= AB + BR + RC + CA
= AB + BD + CQ + CA
[ BP and BR are tangents from point B and CP and CQ are tangents from point C]
= AP + AQ [ AP = AB + BP, AQ = AC + CQ]
= AP + AP [ AP = AQ]
= 2AP
= 2 × 12 [using (i)]
= 24 cm

NCERT Exemplar Solutions Class 10 Maths Chapter 9 Important Topics:

• Several theorems about a tangent to the circle and their proofs.
• Find the number of tangents that can be drawn from any given point
• In this chapter, students will learn that the tangent will be perpendicular to the line joining centre and the point of tangent on the circle.
• Class 10 Maths NCERT exemplar chapter 9 solutions discusses the method that if we draw two tangents on diametrically opposite points of a circle we will observe that these two tangents are parallel.

NCERT Class 10 Exemplar Solutions for Other Subjects:

• NCERT Exemplar Class 10 Maths solutions

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NCERT Class 10 Maths Exemplar Solutions for Other Chapters:

Features of NCERT Exemplar Class 10 Maths Solutions Chapter 9:

These Class 10 Maths NCERT exemplar chapter 9 solutions provide an extension to the learning of circles done in Class 9. The chapter discusses tangents at any point of the circle along with the theorems such as the perpendicular tangent theorem. The chapter on Circles can be better understood and practiced using these NCERT exemplar Class 10 Maths chapter 9 solutions Circles and will be enough to solve other books such as A textbook of Mathematics by Monica Kapoor, NCERT Class 10 Maths, RD Sharma Class 10 Maths, RS Aggarwal Class 10 Maths et cetera.

NCERT exemplar Class 10 Maths solutions chapter 9 pdf download is a free and valuable feature. It provides the students with a pdf download facility to refer to the solutions in an offline environment while studying NCERT exemplar Class 10 Maths chapter 9.

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