R S Aggarwal Solutions for Class 11 Maths Chapter 21 Circles Exercise 21A

R S Aggarwal Solutions Class 11 Maths are given here for Chapter 21 – Circles exercise 21A. These exercise problems are based on the equation of a circle in standard form. The R S Aggarwal study material for Class 11 can help students discover new ways to solve difficult problems. Students can download R S Aggarwal chapter 21 solutions and practice more problems on Circle.

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 rs aggarwal solution class 11 maths chapter 21
 rs aggarwal solution class 11 maths chapter 21
 rs aggarwal solution class 11 maths chapter 21
 rs aggarwal solution class 11 maths chapter 21
 rs aggarwal solution class 11 maths chapter 21
 rs aggarwal solution class 11 maths chapter 21

 

Access Answers of R S Aggarwal Solutions for Class 11 Maths Chapter 21 Circles Exercise 21A Page number 723

Exercise 21A Page No: 723

Question 1: Find the equation of a circle with centre (2, 4) and radius 5.

Solution:

The general form of the equation of a circle is:

(x – h)2 + (y – k)2 = r2 …..(1)

Where, r is the radius of the circle and (h, k) is the centre of the circle.

Here, r = 5, h = 2 and k = 4

Equation (1)⇒

(x – 2)2 + (y – 4)2 = 52

or (x – 2)2 + (y – 4)2 = 25

or x2 + y2 – 4x – 8y – 5 = 0

Which is the required equation.

Question 2: Find the equation of a circle with centre (-3, -2) and radius 6.

Solution:

The general form of the equation of a circle is:

(x – h)2 + (y – k)2 = r2 …..(1)

Where, r is the radius of the circle and (h, k) is the centre of the circle.

Here, r = 6, h = -3 and k = -2

Equation (1)⇒

(x + 3)2 + (y + 2)2 = 62

or (x + 3)2 + (y + 2)2 = 36

or x2 + y2 + 6x + 4y – 23 = 0

Which is the required equation.

Question 3: Find the equation of a circle with centre (a, a) and radius √2.

Solution:

The general form of the equation of a circle is:

(x – h)2 + (y – k)2 = r2 …..(1)

Where, r is the radius of the circle and (h, k) is the centre of the circle.

Here, r = √2, h = a and k = a

Equation (1)⇒

(x – a)2 + (y – a)2 = (√2)2

or (x – a)2 + (y – a)2 = 2

or x2 + y2 – 2ax – 2ay + (2a2 – 2) = 0

Which is the required equation.

Question 4: Find the equation of a circle with centre (a cos ∝, a sin ∝) and radius a

Solution:

The general form of the equation of a circle is:

(x – h)2 + (y – k)2 = r2 …..(1)

Where, r is the radius of the circle and (h, k) is the centre of the circle.

Here, r = a, h = a cos ∝ and k = a sin ∝

Equation (1)⇒

(x – a cos ∝)2 + (y – a sin ∝)2 = (a)2

or (x – a cos ∝)2 + (y – a sin ∝)2 = a2

or x2 + y2 – 2a cos ∝ x – 2a sin ∝ y + a2 cos2 ∝ + a2 sin2 ∝ = a2

or x2 + y2 – 2a cos ∝ x – 2a sin ∝ y = 0

[Because cos2 ∝ + a2 sin2 ∝ = 1]

Which is the required equation.

Question 5: Find the equation of a circle with centre (-a, -b) and radius √(a2 – b2).

Solution:

The general form of the equation of a circle is:

(x – h)2 + (y – k)2 = r2 …..(1)

Where, r is the radius of the circle and (h, k) is the centre of the circle.

Here, r = √(a2 – b2), h = -a and k = -b

Equation (1)⇒

(x + a )2 + (y + b)2 = (√(a2 – b2))2

or (x + a )2 + (y + b)2 = a2 – b2

or x2 + y2 + 2a x + 2ay + a2 + b2 = a2 – b2

or x2 + y2 + 2a x + 2ay + 2 b2 = 0

Which is the required equation.

Question 6: Find the equation of a circle with centre at the origin and radius 4.

Solution:

The general form of the equation of a circle is:

(x – h)2 + (y – k)2 = r2 …..(1)

Where, r is the radius of the circle and (h, k) is the centre of the circle.

Here, r = 4, h = 0 and k = 0

Equation (1)⇒

(x – 0 )2 + (y – 0)2 = (4)2

or x2 + y2 = 16

or x2 + y2 – 16 = 0

Which is the required equation.

Question 7: Find the centre and radius of each of the following circles:

(i) (x – 3)2 + (y – 1)2 = 9

(ii)

R S Agrawal Solution Class 11 chapter 21 question 7

(iii) (x + 5)2 + (y – 3) 2 = 20

(iv) x2 + (y – 1) 2 = 2

Solution:

(i) The general form of the equation of a circle is:

(x – h)2 + (y – k) 2 = r2

Where, (h, k) is the centre of the circle.

r = radius of the circle.

Given equation is (x – 3)2 + (y – 1)2 = 9

Comparing the given equation of circle with general form we get:

h = 3 , k = 1, r2 = 9

Centre = (3, 1) and radius = 3 units.

(ii) The general form of the equation of a circle is:

(x – h)2 + (y – k) 2 = r2

Where, (h, k) is the centre of the circle.

r = radius of the circle.

Given equation is

R S Agrawal Solution Class 11 chapter 21 question 7 solution

Comparing the given equation of circle with general form we get:

h = 1/2 , k = – 1/3, r2 = 1/16

So, Centre = (1/2, – 1/3) and radius = 1/4 units.

(iii) The general form of the equation of a circle is:

(x – h)2 + (y – k) 2 = r2

Where, (h, k) is the centre of the circle.

r = radius of the circle.

Given equation is

(x + 5)2 + (y – 3) 2 = 20

Comparing the given equation of circle with general form we get:

h = – 5 , k = 3, r2 = 20

Centre = ( -5, 3) and radius = √20 or 2√5 units.

(iv) The general form of the equation of a circle is:

(x – h)2 + (y – k) 2 = r2

Where, (h, k) is the centre of the circle.

r = radius of the circle.

Given equation is

x2 + (y – 1) 2 = 2

Comparing the given equation of circle with general form we get:

h = 0 , k = 1, r2 = 2

So, Centre = (0, 1) and radius = √2units.

Question 8: Find the equation of the circle whose centre is (2, – 5) and which passes through the point (3, 2).

Solution:

The general form of the equation of a circle is:

(x – h)2 + (y – k) 2 = r2 …..(1)

Where, (h, k) is the centre of the circle.

r = radius of the circle.

We are given with, centre = (2, – 5)

Or (h, k) = (2, – 5)

Find the radius of circle:

Since the circle passes through (3, 2), so it must satisfy the equation.

Put x = 3 and y = 2 in (1)

(3 – 2)2 + (2 + 5) 2 = r2

1 + 49 = r2

Or r2 = 50

Now,

Equation of circle is:

(x – 2)2 + (y + 5)2 = 50

Which is required equation.


Access other exercise solutions of Class 11 Maths Chapter 21 Circles

Exercise 21B Solutions

R S Aggarwal Solutions for Class 11 Maths Chapter 21 Exercise 21A

Class 11 Maths Chapter 21 Circles Exercise 21A is based on the following topics :

  • Equation of a circle in standard form
  • Equation of a circle, the end points of whose diameter are given