NCERT Solutions for Class 10 Maths Chapter 10 Circles

NCERT
SOLUTIONS
CLASS - 10th

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Class : 10th
Subject : Maths
Chapter : 10
Chapter Name : Circles

Exercise 10.1

Q1 How many tangents can a circle have?

Answer. Tangent is a line that intersects the circle at one point There are in nite number of
points on circle At every point, there is one tangent
Hence, there are in nite number of tangents in a circle.

Page : 209 , Block Name : Exercise 10.1

Q2 Fill in the blanks :
(i) A tangent to a circle intersects it in point (s).
(ii) A line intersecting a circle in two points is called a .
(iii) A circle can have parallel tangents at the most.
(iv) The common point of a tangent to a circle and the circle is called .

Answer. (i)

Note only there can be one tangent at point P i.e. tangent XY If we try to make more than one
line at point P example AB,it becomes a secant (as it intersects at more than one point)
(ii)

(iii)

Note only two tangents are possible AB and XYCD is not a tangent it is a secant as it intersects
at 2 points
(iv)

Here P is the point of contact.

Page : 209 , Block Name : Exercise 10.1

Q3 A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a
point Q so that OQ = 12 cm. Length PQ is :
(A) 12 cm

(B) 13 cm
(C) 8.5 cm
(D) √119cm

Answer. Given OP radius 5 cm & OQ = 12 cm

Since PQ is a tangent,
OP ⊥ PQ
∘
So, ∠OPQ = 90

Hence, ΔOPQ is a right triangle

ΔOPQ

Using Pythagoras theorem
2 2 2
(Hypotenuse ) = ( Height ) + ( Base )

2 2 2
OQ = (OP) + (PQ)

2 2 2
12 = 5 + (PQ)

2
144 = 25 + (PQ)

2
144 − 25 = PQ

PQ = √119

So (D) is correct

Page : 209 , Block Name : Exercise 10.1

Exercise 10.2

Q1 From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from
the centre is 25 cm. The radius of the circle is
(A) 7 cm
(B) 12 cm
(C) 15 cm

(D) 24.5 cm

Answer.

Let O be the centre of the circle.
Given that,
OQ=25cm and PQ=24 cm
As the radius is perpendicular to the tangent at the point the contact
Therefore ,OP ⊥ P Q
Applying pythagoras theorem in triangle OPQ, we obtain
2 2 2
OP + PQ = OQ
2 2 2
OP + 24 = 25
2
OP = 625 − 576
2
OP = 49

OP=7
Therefore , the radius of circle is 7 cm.
Hence ,alternative (A) is correct.

Q2 In Fig. 10.11, if TP and TQ are the two tangents to a circle with centre O so that ∠PTQ=
110 , then ∠PTQ is equal to
∘

(A) 60 ∘

(B) 70 ∘

(C) 80 ∘

(D) 90 ∘

Answer. It is given that TP and TQ are tangents.
Therefore, radius drawn to these tangents will be perpendicular to the tangents.
Thus, OP ⊥ T P and OQ ⊥ T O
Angle OPT= 90 ∘

Angle OQT = 90 ∘

In quadrilateral POQT,
Sum of all interior angles = 360 ∘

∘
∠OP T + ∠P OQ + ∠OQT + ∠P T Q = 360
∘ ∘ ∘ ∘
90 + 110 + 90 + ∠P T Q = 360
∘
∠PTQ = 70

Hence ,alternative (B) is correct.

Q3 If tangents PA and PB from a point P to a circle with centre O are inclined to each other at
angle of 80 , then ∠POA is equal to
∘

(A) 50 ∘

(B) 60 ∘

(C) 70 ∘

(D) 80 ∘

Answer. It is given that PA and PB are tangents

Therefore, the radius drawn to these tangents will be perpendicular to the tangents.
Thus, OA ⊥ P A and 0B ⊥ P B
∘
∠OBP = 90
∘
∠OAP = 90

In AOBP,
Sum of all interior angles = 360 ∘

∘
∠OAP + ∠AP B + ∠P BO + ∠BOA = 360
∘ ∘ ∘ ∘
90 + 80 + 90 + ∠BOA = 360
∘
∠BOA = 100

△OP B and ΔOP A′

AP=BP( tangents from a point)
OA=OB(radii of the circle)
OP=OP(common side)
Therefore ,△OP B ≈ △OP A ( SSS congruence criterion)
A ↦ B, P → P , O → O

And thus, ∠POB = ∠POA
∘
1 100 ∘
∠POA = ∠AOB = = 50
2 2

Hence , alternative (A) is correct.

Q4 Prove that the tangents drawn at the ends of a diameter of a circle are parallel.

Answer.

Let AB be a diameter of the circle. Two tangents PQ and RS are drawn at points A
and B respectively.
Radius drawn to these tangents will be perpendiculæ to the tangents.
Thus, OA ⊥ RS and 0B ⊥ P Q
∘
∠OAR = 90
∘
∠OAS = 90
∘
∠OBP = 90
∘
∠OBQ = 90

I can be observed that ,
∠OAR = ∠OBQ (Alternate interior angle)

∠OAS = ∠OBP (Alternate interior angle)

Since alternate interior angles equal, lines PQ and RS will be parallel.

Q5 Prove that the perpendicular at the point of contact to the tangent to a circle passes
through the centre.

Answer. Let us consider a circle With centre O. Let AB be a tangent which touches the circle at
P.

We have to prove that the line to AB at p passes through centre O. We shallprove this by
contradiction method.

Let us assume that the perpendicular to AB at P does not pass through centre O. Let it pass
through another point O'. Join OP and O'P.

As perpendicular to AB at P passes through O', therefore,
∠O P B = 90 ......(1)
′ ∘

O is tie centre of the circle and P is the point of contact. We know the line joining
the centre and the point Of contact to the tangent Of the circle are perpendicular to
each other.
∠OP B = 90 ................(2)
∘

Comparing equations (1) and (2), we obtain
∠O P B = ∠OP B..............(3)
′

From the gure, it can be observed that,
∠O P B < ∠OP B...............(4)
′

Therefore , ∠O P B < ∠OP B is not possible .it is only possible , when the line O P coincides
′ ′

with OP.
Therefore the perpendicular to AB through P passes through centre O.

Q6 The length of a tangent from a point A at distance 5 cm from the centre of the circle is 4
cm. Find the radius of the circle.

Answer.

Let us consider a circle at point O.
AB is a tangent drawn on this circle from point A.
Given that,
OA = 5cm and AB = 4 cm
In triangle ABO,
OB ⊥ AB

Applying the pythagoras theorem in triangle ABO , we obtain
2 2 2
AB + BO = OA
2 2 2
4 + BO = 5

2
16 + BO = 25
2
BO = 9

BO=3 cm
Hence , the radius of the circle is 3 cm

Q7 Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger
circle which touches the smaller circle.

Answer.

Let the two concentric circles be centered at point O. And let PQ be the chord of the
larger circle which touches the smaller circle at point A, Therefore, PQ is tangent to
the smaller- circle,
OA ⊥ P Q (As OA is the radius Of the circle)

Applying Pythagoras theorem in triangle OAP, we obtain
2 2 2
OA + AP = OP
2 2 2
3 + AP = 5
2
9 + AP = 25
2
AP = 16

AP=4
Since OA ⊥ P Q
AP = AQ (Perpendicular from the center Of the circle bisects the chord)
∴ P Q = 2AP = 2 × 4 = 8

Therefore, the length of the chord of the larger circle is 8 cm.

Q8 A quadrilateral ABCD is drawn to circumscribe a circle (see Fig. 10.12). Prove that AB + CD
= AD + BC.

Answer. It can be observed that
DR = DS (Tangents on the circle from point D) …………… (1)
CR = CQ (Tangents on the circle from point C) …………… (2)
BP = BQ (Tangents on the circle from point B) …………… (3)
AP = AS on the circle from point A) ……………………… (4)
Adding all these equations, we obtain
DR + CR + BP + AP = DS + CQ + BQ + AS

(DR + CR) + (BP + AP ) = (DS + AS) + (CQ + BQ)

CD + AB = AD + BC

Q9 In Fig. 10.13, XY and X Y are two parallel tangents to a circle with centre O and another
′ ′

tangent AB with point of contact C intersecting XY at A and X Y at B. Prove that ∠AOB=
′ ′

90 .
∘

Answer. Let us joint point O to C.

△OPA and △OCAr

OP = OC (Radii of the same circle)
AP=AC (Tangents from point A)
AO = AO (Common side)

ΔOP A ≅ △OCA (SSS congruence criterion)
Therefore, P ↦ C, A ↦ A, O ↦ O
∠POA = ∠COA..................(1)

Similarly △OQB ≅ △OCB
∠QOB = (∠COS .......................(2)

Since P O Q is a diameter of the circle, it is a straight line.
Therefore ∠P OA + ∠COA + ∠COB + ∠QOB = 180 ∘

From equations (1) and (2), it can be observed that
∘
2∠COA + 2∠COB = 180
∘
∠COA + ∠COB = 90
∘
∠AOB = 90

Q10 Prove that the angle between the two tangents drawn from an external point to a circle is
supplementary to the angle subtended by the line-segment joining the points of contact at the
centre.

Answer.

Let us consider a circle centered at point O. Let P be external point from which two tangents
PA and PB are drawn to the circle which are touching the circle at point
A and B respectively and AB is the line segment, joining point Of contacts A and 3 together
such that it subtends Angle AOB at centre- O of the circle.
It can be observed that
OA(radius) ⊥ P A(tangent)

Therefore ,∠OAP = 90 ∘

Similarly , OB ⊥ P B(tangent)
∘
∠OBP = 90

In quadrilateral OAPB,
Sum of the all interior angles
∘
∠OAP + ∠AP B + ∠P BO + ∠BOA = 360
∘ ∘
900 + ∠AP B + 90 + ∠BOA = 360
∘
∠AP B + ∠BOA = 180

Hence, it can be observed that the angle between the two tangents drawn from an
external point to a circle is supplementary to the angle subtended by the line-
segment joining the points of contact at the centre.

Q11 Prove that the parallelogram circumscribing a circle is a rhombus.

Answer. Since ABCO is a parallelogram,
AB=AD…………(1)
BC=AD…………(2)

It can be observed that
DR = DS (Tangents on the circle from point D)
CR = CQ (Tangents on the circle from point C)
BP=BQ (Tangents on the circle from point B)
AP= AS (Tangents on the circle from point A)
Adding all these equations, we obtain
OR + CR + BP + AP DS + CQ + BQ + AS
(DR + CR) + (BP + AD) = (DS + AS) + (CQ + BQ)
CD + AB = AD + BC
On putting the values of equations (1 ) and (2) in this equation, we obtain
2A8 =2BC
AB=BC……….(3)
Comparing equations (1), (2), (3), we obtain
AS = BC = CD = DA
Hence, ABCD is a rhombus.

Q12 A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD
and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm
respectively (see Fig. 10.14). Find the sides AB and AC.

Answer.

Let the given circle touch the sides AB and AC of the triangle at point E and F respectively and
the Length of the line segment AF be x.
In triangle ABC,
CF=CD =6cm (Tangents on the circle from point C)
BE= BD = 8cm (Tangents on the circle from point B)
AE=AF =x cm (Tangents on the circle from point A)
AB=AE+EB=x + 8
BC=BD+ DC = 8+6=14
CA=CF+FA=6+x
2s=AB + BC + CA
=x+8+14+6+x
=28+2x
s=14+x
ΔABC = √s(s − a)(s − b)(s − c)

= √{14 + x}{(14 + x) − 14}{(14 + x) − (6 + x)}(14 + x) − (8 + x)}

= √(14 + x)(x)(8)(6)

2
= 4√3 (14x + x )

Area of ΔOBC = 1

2
× OD × BC =
1

2
× 4 × 14 = 28

Area of ΔOCA = 1

2
× OF × AC =
1

2
× 4 × (6 + x) = 12 + 2x

Area of 1

2
× OE × AB =
1

2
× 4 × (8 + x) = 16 + 2x

ΔABC = Area of ΔOBC + Area of ΔOCA + Area of △OAB
2
4√3 (14x + x ) = 28 + 12 + 2x + 16 + 2x

2
⇒ 4√3 (14x + x ) = 56 + 4x

2
⇒ √3 (14x + x ) = 14 + x
2 2
⇒ 3 (14x + x ) = (14 + x)
2 2
⇒ 42x + 3x = 196 + x + 28x
2
⇒ 2x + 14x − 196 = 0
2
⇒ x + 7x − 98 = 0
2
⇒ x + 14x − 7x − 98 = 0

⇒ x(x + 14) − 7(x + 14) = 0

⇒ (x + 14)(x − 7) = 0

Either x+14=0 or x-7=0

Therefore ,x = -1 and 7
However, x = -14 is not possible as the length of the sides will be negative.
Therefore, x=7
Hence, AB =x+8 = 7+8=15 cm
CA = 6 + x=6 + 7 =13 cm

Q13 Prove that opposite sides of a quadrilateral circumscribing a circle subtend
supplementary angles at the centre of the circle.

Answer. Let ABCD be a quadrilateral circumscribing a circle centered at O such that it touches
the circle at point P, Q, R, S. Let us join the v«tices of the quadrilateral ABCD to the of the
circle.
Consider triangle OAP and triangle OAS,
AP = AS (Tangents from the sarne point)
OP=OS (Radii Of the sane circle)
OA=OA (Common side)
△OAP = △OAS ( SSS congruence criterion)

Therefore ,A <-> A,P <-> S ,O <-> O
And thus ∠POA = ∠AOS
∠1 = ∠8,

Similarly,
∠2 = ∠3,

∠4 = ∠5,

∠6 = ∠7,
∘
∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 + ∠8 = 360
∘
(∠1 + ∠8) + (∠2 + ∠3) + (∠4 + ∠5) + (∠6 + ∠7) = 360
∘
2∠1 + 2∠2 + 2∠5 + 2∠6 = 360
∘
2(∠1 + ∠2) + 2(∠5 + ∠6) = 360
∘
(∠1 + ∠2) + (∠5 + ∠6) = 180

Similarly , we can prove that ∠BOC + ∠DOA = 180 ∘

Hence, opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles
at the centre of the circle .