NCERT Solutions for Class 10 Maths Chapter 10 : Circles

Q: Q1. What are the most important topics in NCERT Class 10 Chapter 10 Circles?

Answer: NCERT Solutions for Class 10 Maths Chapter 10 covers the following topics: introduction to circles, tangent to a circle, lengths of tangents, number of tangents from a point on a circle, the radius-tangent relationship, and the chapter summary.

Q: Q2. How many exercises are there in Chapter 10 of Class 10 Maths?

Answer: NCERT Solutions Class 10 Maths Chapter 10 Circles contains a total of 17 questions separated into 2 exercises. Class 10 Maths Chapter 10 Circles contains a total of 17 sums, 10 of which are simple, 3 of which are fairly complicated, and 4 of which are tough.

Q: Q3. How many tangents can a circle have?

Answer: Generally, there is no limit to the number of tangents a circle can have. It can be limitless since a circle is made up of infinite points that are all equally far from one another. Because there are infinite locations on the circumference of a circle, infinite tangents can be traced from them.

Q: Q4. Do I have to practice all of the questions in NCERT Solutions Class 10 Maths Chapter 10 Circles?

Answer: All questions in NCERT Solutions Class 10 Maths Circles are based on key concepts that provide a comprehensive overview of the subject. Because the problems can be complex, practising all of the sums, as well as the examples, can help students learn the idea of circles and understand how to use various formulas. Students must also revise the given theorems and practice questions twice because the sums are proof-based.

Question 1

How many tangents can be drawn at any point on the circle?

A

0

B

1

C

2

D

3

Medium

Solution

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Answer

Let P be any point on the circle with center $O$

$O P = r a d i u s$

Take a line L through P and Q as shown if L is perpendicular to OP

$O Q ⊥ O P$ because the perpendicular distance is shortest.

Every point except P lies outside the circle and line $l$ must be a tangent. At any given point one and only one tangent can be drawn.

Answer

Let P be any point on the circle with center $O$

$O P = r a d i u s$

Take a line L through P and Q as shown if L is perpendicular to OP

$O Q ⊥ O P$ because the perpendicular distance is shortest.

Every point except P lies outside the circle and line $l$ must be a tangent. At any given point one and only one tangent can be drawn.

Question 2

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) A common point of a tangent to a circle and the circle is called ______.

Easy

Solution

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Answer

(i) One point

Circle is the locus of points equidistant from a given point, the center of the circle, and nd Tangent is the line which intersect circle at one point only

(ii) Secant

A line which intersect circle in two points is called a chord if this line passes through center then it is called secant. (Points are A and B)

(iii) Two

As circle has infinite points, so there will be infinite tangents can be drawn on these points which touches at only one point. (P and Q)

So there will be infinite pairs of tangents which are parallel.

(iv) Point of Contact

The common point of a tangent to a circle and the circle is called point of contact. (P or Q in the given figure)

Answer

(i) One point

Circle is the locus of points equidistant from a given point, the center of the circle, and nd Tangent is the line which intersect circle at one point only

(ii) Secant

A line which intersect circle in two points is called a chord if this line passes through center then it is called secant. (Points are A and B)

(iii) Two

As circle has infinite points, so there will be infinite tangents can be drawn on these points which touches at only one point. (P and Q)

So there will be infinite pairs of tangents which are parallel.

(iv) Point of Contact

The common point of a tangent to a circle and the circle is called point of contact. (P or Q in the given figure)

Question 3

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

O Q = 12

cm. Length of PQ is :

A

112

cm

B

113

cm

C

85

cm

D

119

cm

Medium

Solution

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Answer

Given,

Radius OP=

5

cm and OQ=

12

cm

PQ is the tangent to the circle.

∠ O P Q = 9 0^{0}

So,by Pythagoras theorem we get,

P Q^{2} = O Q^{2} - O P^{2}

= > P Q^{2} = 1 2^{2} - 5^{2}

= > P Q^{2} = 144 - 25

= > P Q^{2} = 119

= > P Q = 119

cm

Answer

Given,

Radius OP=

5

cm and OQ=

12

cm

PQ is the tangent to the circle.

∠ O P Q = 9 0^{0}

So,by Pythagoras theorem we get,

P Q^{2} = O Q^{2} - O P^{2}

= > P Q^{2} = 1 2^{2} - 5^{2}

= > P Q^{2} = 144 - 25

= > P Q^{2} = 119

= > P Q = 119

cm

Question 4

Draw a circle and two lines parallel to a given line that one is a tangent and the other, a secant to the circle

Medium

Solution

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Answer

Let the circle has a center

O

and

A B

be a given line such that

Let

A B ∥ C D ∥ E F

.

Where

C D

is a second ans

E F

is a tangent intersecting the circle at

R

.

Answer

Let the circle has a center

O

and

A B

be a given line such that

Let

A B ∥ C D ∥ E F

.

Where

C D

is a second ans

E F

is a tangent intersecting the circle at

R

.

Question 5

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

Medium

Solution

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Answer

Tangent=

x y

point of contact=

b

length of the tangent to a circle =

24 c m

i.e.,

P Q = 24 c m

O Q = 25 c m

prove

radius of circle i.e., OP

Proof

Since

x y

is tangent

O P ⊥ X Y

so,

∠ O P Q = 90

So,

△ O P Q

is a right angled triangle using by pythagoras

theorem

O Q^{2} = O P^{2} + P Q^{2}

= > 2 5^{2} = O P^{2} + 2 4^{2}

= > O P^{2} = 625 - 576

= > O P^{2} = 49

= > O P = 7

Hence ,

radius=

O P = 7 c m

Answer

Tangent=

x y

point of contact=

b

length of the tangent to a circle =

24 c m

i.e.,

P Q = 24 c m

O Q = 25 c m

prove

radius of circle i.e., OP

Proof

Since

x y

is tangent

O P ⊥ X Y

so,

∠ O P Q = 90

So,

△ O P Q

is a right angled triangle using by pythagoras

theorem

O Q^{2} = O P^{2} + P Q^{2}

= > 2 5^{2} = O P^{2} + 2 4^{2}

= > O P^{2} = 625 - 576

= > O P^{2} = 49

= > O P = 7

Hence ,

radius=

O P = 7 c m

Question 6

If

T P

and

T Q

are two tangents to a circle with centre

O

so that

∠ P O Q = 110^{o}

, then

∠ P T Q

is equal to

A

60^{o}

B

70^{o}

C

80^{o}

D

90^{o}

Medium

Solution

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Answer

Given,

∠ P O Q = 11 0^{\circ}

We know,

∠ O P T = ∠ O Q T = 9 0^{\circ}

(Angle between the tangent and the radial line at the point of intersection of the tangent at the circle)
Now, in quadrilateral

P O Q T

Sum of angles

= 360

∠ O P T + ∠ O Q T + ∠ P T Q + ∠ P O Q = 36 0^{\circ}

90 + 90 + ∠ P T Q + 110 = 360

∠ P T Q = 360 - 290

∠ P T Q = 7 0^{\circ}

Answer

Given,

∠ P O Q = 11 0^{\circ}

We know,

∠ O P T = ∠ O Q T = 9 0^{\circ}

(Angle between the tangent and the radial line at the point of intersection of the tangent at the circle)
Now, in quadrilateral

P O Q T

Sum of angles

= 360

∠ O P T + ∠ O Q T + ∠ P T Q + ∠ P O Q = 36 0^{\circ}

90 + 90 + ∠ P T Q + 110 = 360

∠ P T Q = 360 - 290

∠ P T Q = 7 0^{\circ}

Question 7

If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of

8 0^{\circ}

, then

∠ P O A

is equal to:

Medium

Solution

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Answer

Given that,

PA and PB are two tangents a circle and $∠ A P B = 80^{0}$

To find that $∠ P O A = ?$

Construction:- join $O A, O B a n d O P$

Proof:- Since $O A ⊥ P A$ and $O B ⊥ P B$

Then $∠ O A P = 90^{0}$ and $∠ O B P = 90^{0}$

In

$Δ O A P & Δ O B P$

$O A = O B (r a d i u s)$

$O P = O P (C o m m o n)$

$P A = P B (l e n g t h s o f tan g e n t d r a w n f r o m e x t e r n a l p o i n t i s e q u a l)$

$∴ Δ O A P ≅ Δ O B P (S S S c o n g r u e n c y)$

So,

$[∠ O P A = ∠ O P B (b y C P C T)]$

So,

$∠ O P A = \frac{1}{2} ∠ A P B$

$= \frac{1}{2} \times 80^{0} = 40^{0}$

In $Δ O P A,$

$∠ P O A + ∠ O P A + ∠ O A P = 180^{0}$

$∠ P O A + 40^{0} + 90^{0} = 180^{0}$

$∠ P O A + 130^{0} = 180^{0}$

$∠ P O A = 180^{0} - 130^{0}$

$∠ P O A = 50^{0}$

The value of $∠ P O A$ is $50^{0} .$

Answer

Given that,

PA and PB are two tangents a circle and $∠ A P B = 80^{0}$

To find that $∠ P O A = ?$

Construction:- join $O A, O B a n d O P$

Proof:- Since $O A ⊥ P A$ and $O B ⊥ P B$

Then $∠ O A P = 90^{0}$ and $∠ O B P = 90^{0}$

In

$Δ O A P & Δ O B P$

$O A = O B (r a d i u s)$

$O P = O P (C o m m o n)$

$P A = P B (l e n g t h s o f tan g e n t d r a w n f r o m e x t e r n a l p o i n t i s e q u a l)$

$∴ Δ O A P ≅ Δ O B P (S S S c o n g r u e n c y)$

So,

$[∠ O P A = ∠ O P B (b y C P C T)]$

So,

$∠ O P A = \frac{1}{2} ∠ A P B$

$= \frac{1}{2} \times 80^{0} = 40^{0}$

In $Δ O P A,$

$∠ P O A + ∠ O P A + ∠ O A P = 180^{0}$

$∠ P O A + 40^{0} + 90^{0} = 180^{0}$

$∠ P O A + 130^{0} = 180^{0}$

$∠ P O A = 180^{0} - 130^{0}$

$∠ P O A = 50^{0}$

The value of $∠ P O A$ is $50^{0} .$

Question 8

Prove that the tangents drawn at the end of a diameter of a circle are parallel

Medium

Solution

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Answer

To prove: PQ

∣ ∣

RS

Given: A circle with centre O and diameter AB. Let PQ be the tangent at point A & Rs be the point B.

Proof: Since PQ is a tangent at point A.

OA

⊥

PQ(Tangent at any point of circle is perpendicular to the radius through point of contact).

∠ O Q P = 9 0^{o}

…………

(1)

OB

⊥

RS

∠ O B S = 9 0^{o}

……………

(2)

From

(1)

&

(2)

∠ O A P = ∠ O B S

i.e.,

∠ B A P = ∠ A B S

for lines PQ & RS and transversal AB

∠ B A P = ∠ A B S

i.e., both alternate angles are equal.

So, lines are parallel.

$$\therefore PQ||RS.

Answer

To prove: PQ

∣ ∣

RS

Given: A circle with centre O and diameter AB. Let PQ be the tangent at point A & Rs be the point B.

Proof: Since PQ is a tangent at point A.

OA

⊥

PQ(Tangent at any point of circle is perpendicular to the radius through point of contact).

∠ O Q P = 9 0^{o}

…………

(1)

OB

⊥

RS

∠ O B S = 9 0^{o}

……………

(2)

From

(1)

&

(2)

∠ O A P = ∠ O B S

i.e.,

∠ B A P = ∠ A B S

for lines PQ & RS and transversal AB

∠ B A P = ∠ A B S

i.e., both alternate angles are equal.

So, lines are parallel.

$$\therefore PQ||RS.

Question 9

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

Medium

Solution

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Answer

Given a circle with center

O

and

A B

the tangent intersecting circle at point

P

and prove that

O P ⊥ A B

We know that tangent of the circle is perpendicular to radius at points of contact Hence

O P ⊥ A B

So,

∠ O P B = 9 0^{o} . . . . . . . . . . (i)

Now lets assume some point

X

Such that

X P ⊥ A N

Hence

∠ X P B = 9 0^{o} . . . . . . . . . (i i)

From eq

(i)

&

(i i)

∠ O P B = ∠ X P B = 9 0^{o}

Which is possible only if line

X P

passes though

O

Hence perpendicular to tangent passes though centre

Answer

Given a circle with center

O

and

A B

the tangent intersecting circle at point

P

and prove that

O P ⊥ A B

We know that tangent of the circle is perpendicular to radius at points of contact Hence

O P ⊥ A B

So,

∠ O P B = 9 0^{o} . . . . . . . . . . (i)

Now lets assume some point

X

Such that

X P ⊥ A N

Hence

∠ X P B = 9 0^{o} . . . . . . . . . (i i)

From eq

(i)

&

(i i)

∠ O P B = ∠ X P B = 9 0^{o}

Which is possible only if line

X P

passes though

O

Hence perpendicular to tangent passes though centre

Question 10

The length of a tangent from a point A at distance

5

cm from the centre of circle is

4

cm. The radius of the circle is

A

1

cm

B

2

cm

C

3

cm

D

4

cm

Medium

Solution

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Answer

Let

A T

be the tangent drawn from a point

A

to a circle with centre

O

and

O A = 5

cm and

A T = 4

cm. Since tangent at a point is

perpendicular to the radius through the point of contact

∴ O T ⊥ A T

∴

from right angled

△ O A T

,

(O A)^{2} = (O T)^{2} + (T A)^{2}

\Rightarrow (5)^{2} = (O T)^{2} + (4)^{2}

\Rightarrow 25 - 16 = (O T)^{2}

\Rightarrow 9 = (O T)^{2}

\Rightarrow O T = 3

cm

∴

radius of the circle

= 3

cm.

Answer

Let

A T

be the tangent drawn from a point

A

to a circle with centre

O

and

O A = 5

cm and

A T = 4

cm. Since tangent at a point is

perpendicular to the radius through the point of contact

∴ O T ⊥ A T

∴

from right angled

△ O A T

,

(O A)^{2} = (O T)^{2} + (T A)^{2}

\Rightarrow (5)^{2} = (O T)^{2} + (4)^{2}

\Rightarrow 25 - 16 = (O T)^{2}

\Rightarrow 9 = (O T)^{2}

\Rightarrow O T = 3

cm

∴

radius of the circle

= 3

cm.

NCERT Solutions for Class 10 Maths Chapter 10 : Circles

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NCERT Solutions for Class 10 Maths Chapter 10 : Circles

NCERT Solutions for Class 10 Maths Chapter 10 Circles – Brief Overview

1. Circles

2. Tangent to a Circle

3. Number of Tangents From a Point

Frequently Asked Questions on NCERT Class 10 Maths Chapter 10 : Circles