R S Aggarwal Solutions for Class 11 Maths Chapter 21 Circles

R S Aggarwal Class 11 Maths Solutions for Chapter 11 are provided here. In this chapter, students will learn how to find the equation of a circle under different conditions.

Important Points to Remember:

1. The general form of the equation of a circle : (x – h)2 + (y – k)2 = r2

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

2. The equation of a circle with centre at the origin and radius r is given by

x2 + y2 = r2

3. The equation of a circle with A(x1 , y1 ) and B(x2 , y2) as the end points of a diameter is given by (x – x1)(x – x2) + (y – y1)(y – y2) = 0

Check the detailed R S Aggarwal Solutions for all Class 11 chapters and start practicing to score good marks.

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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
 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
 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 to Maths R S Aggarwal Chapter 21 Circles

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.


Exercise 21B Page No: 730

Question 1: Show that the equation x2 + y2 – 4x + 6y – 5 = 0 represents a circle. Find its centre and radius.

Solution:

Given equation is x2 + y2 – 4x + 6y – 5 = 0

The general equation of a circle is as follows:

x2 + y2 + 2gx + 2fy + c = 0

Where g, f and c are constants

With

Centre: (-g, -f)

R S Aggarwal Solution Class 11 chapter 21 ex 21B question 1

On comparing given equation with general form of circle, we have

2g = – 4 ⇒ g = -2

2f = 6 ⇒ f = 3 and

c = – 5

Centre: (-g, -f) = (2, -3)

R S Aggarwal Solution Class 11 chapter 21 ex 21B question 1 solution

Question 2: Show that the equation x2 + y2 + x – y = 0 represents a circle. Find its centre and radius.

Solution:

Given equation is x2 + y2 + x – y = 0

The general equation of a circle is as follows:

x2 + y2 + 2gx + 2fy + c = 0

Where g, f and c are constants

With

Centre: (-g, -f)

R S Aggarwal Solution Class 11 chapter 21 ex 21B question 2

On comparing given equation with general form of circle, we have

2g = 1 ⇒ g = 1/2

2f = – 1 ⇒ f = -1/2 and

c = 0

Now,

Centre: (-g, -f) = (-1/2, 1/2)

R S Aggarwal Solution Class 11 chapter 21 ex 21B question 2 solution

Question 3: Show that the equation 3x2 + 3y2 + 6x – 4y – 1 = 0 represents a circle. Find its centre and radius.

Solution:

Given equation is 3x2 + 3y2 + 6x – 4y – 1 = 0

Or x2 + y2 + 2x – 4/3 y – 1/3 = 0

The general equation of a circle is as follows:

x2 + y2 + 2gx + 2fy + c = 0

Where g, f and c are constants.

With

Centre: (-g, -f)

R S Aggarwal Solution Class 11 chapter 21 ex 21B question 3

On comparing given equation with general form of circle, we have

2g = 2 ⇒ g = 1

2f = -4/3 ⇒ f = -2/3 and

c = -1/3

Now,

Centre (-g, -f) = (-1, 2/3)

R S Aggarwal Solution Class 11 chapter 21 ex 21B question 3 solution

Question 4: Show that the equation x2 + y2 + 2x + 10y + 26 = 0 represents a point circle. Also, find its centre.

Solution:

Given equation is x2 + y2 + 2x + 10y + 26 = 0

The general equation of a circle:

x2 + y2 + 2gx + 2fy + c = 0

where c, g, f are constants.

On comparing given equation with general equation of circle, we have

2g = 2 ⇒ g = 1

2f = 10 ⇒ f = 5 and

c = 26

Now,

Centre ( – g, – f) = ( – 1, – 5).

R S Aggarwal Solution Class 11 chapter 21 ex 21B question 4

Since radius is zero, Thus it is a point circle with radius zero.

Question 5: Show that the equation x2 + y2 – 3x + 3y + 10 = 0 does not represent a circle.

Solution:

Given equation is x2 + y2 – 3x + 3y + 10 = 0

We know that, any equation with negative radius(complex number) does not represent a circle.

Find radius of given equation:

R S Aggarwal Solution Class 11 chapter 21 ex 21B question 5

Radius is a complex number. Therefore, given equation does not represent a circle.

Question 6: Find the equation of the circle passing through the points

(i) (0, 0), (5, 0) and (3, 3)

(ii) (1, 2), (3, – 4) and (5, – 6)

(iii) (20, 3), (19, 8) and (2, – 9)

Also, find the centre and radius in each case.

Solution:

Before we start solving listed problems, students are advised to keep below information in mind.

The general equation of a circle is as follows:

x2 + y2 + 2gx + 2fy + c = 0

Where g, f and c are constants.

With

Centre: (-g, -f)

R S Aggarwal Solution Class 11 chapter 21 ex 21B question 6

(i) (0, 0), (5, 0) and (3, 3)

The Circle equation is:

R S Aggarwal Solution Class 11 chapter 21 ex 21B question 6 part 1

Let us apply Laplace Expansion to solve this problem:

R S Aggarwal Solution Class 11 chapter 21 ex 21B question 6 part 1 solution

On comparing above equation with the general form of circle, we get

2g = -5 ⇒ g = -2.5

2f = -1 ⇒ f = -0.5

c = 0

Now,

centre = (2.5, 0.5)

R S Aggarwal Solution Class 11 chapter 21 ex 21B question 6 part 1 solutions

(ii) (1, 2), (3, – 4) and (5, – 6)

The Circle equation is:

R S Aggarwal Solution Class 11 chapter 21 ex 21B question 6 part ii solution

Let us apply Laplace Expansion to solve this problem:

R S Aggarwal Solution Class 11 chapter 21 ex 21B question 6 part 2 solution

On comparing above equation with the general form of circle, we get

2g = -22 ⇒ g = -11

2f = -4 ⇒ f = -2

c = -25

Now,

Centre = (11, 2)

R S Aggarwal Solution Class 11 chapter 21 ex 21B question 6 part ii solutions

(iii) (20, 3), (19, 8) and (2, – 9)

The Circle equation is:

R S Aggarwal Solution Class 11 chapter 21 ex 21B question 6 part 3

Let us apply Laplace Expansion to solve this problem:

R S Aggarwal Solution Class 11 chapter 21 ex 21B question 6 part iii solution

On comparing above equation with the general form of circle, we get

2g = -14 ⇒ g = -7

2f = -6 ⇒ f = -3

c = -111

Now,

Centre = (7, 3)

R S Aggarwal Solution Class 11 chapter 21 ex 21B question 6 part  3 solution

Question 7: Find the equation of the circle which is circumscribed about the triangle whose vertices are A( – 2, 3), b(5, 2) and C(6, – 1). Find the centre and radius of this circle.

Solution:

Since circle is circumscribed about the triangle whose vertices are A( – 2, 3), B(5, 2) and C(6, – 1), which implies points A, B and C are lie on circumference of circle and satisfy its equation.

The general equation of a circle: (x – h)2 + (y – k) 2 = r2 …(i)

where (h, k) is the centre and r is the radius.

Putting A(-2, 3), B(5, 2) and C(6, -1) in above equation, we get

h2 + k2 + 4h – 6k + 13 = r2 ……….(ii)

h2 + k2 – 10h – 4k + 29 = r2 ………(iii)

h2 + k2 – 12h + 2k + 37 = r2 ………(iv)

Subtract (ii) from (iii)

– 14h + 2k + 16 = 0

or – 7h + k + 8 = 0 ….(v)

Subtract (ii) from (iv)

– 16h + 8k + 24 = 0

or -2h + k + 3 = 0 ……(vi)

Solving (v) and (vi), we have

h = 1

(vi) ⇒ -2 x 1 + k + 3 = 0

⇒ k= -1

Therefore,

Centre = (1, – 1)

And,

Equation (ii) ⇒ r = 5

[using values of h and k]

Thus, required equation of the circle is

(x – 1) 2 + (y + 1) 2 = 52

(x – 1) 2 + (y + 1) 2 = 25

Question 8: Find the equation of the circle concentric with the circle x2 + y2 + 4x + 6y + 11 = 0 and passing through the point P(5, 4).

Solution:

Since circles are concentric, which means circles have common centre and different radii.

Equation of given circle, x2 + y2 + 4x + 6y + 11 = 0

The concentric circle will have the equation

x2 + y2 + 4x + 6y + d = 0 ….(1)

As it passes through P(5, 4),

Put x = 5 and y = 4

52 + 42 + 20 + 24 + d= 0

25 + 16 + 20 + 24 + d = 0

d = – 85

Equation (1) ⇒ x2 + y2 + 4x + 6y – 85 = 0

Which is required equation.

Question 9: Show that the points A(1, 0), B(2, – 7), C(8, 1) and D(9, – 6) all lie on the same circle. Find the equation of this circle, its centre and radius.

Solution:

The general equation of a circle: (x – h)2 + (y – k) 2 = r2 …(i)

where (h, k) is the centre and r is the radius.

Consider points (1, 0), (2, – 7) and (8, 1) lie on the circle.

Putting (1, 0), (2, – 7) and (8, 1) in (i)

Putting (1, 0) ⇒ h2 + k2 + 1 – 2h = r2 ……(ii)

Putting (2, – 7) ⇒ h2 + k2 + 53 – 4h + 14k = r2 …….(iii)

Putting (8, 1) ⇒ (8 – h)2 + (1 – k) 2 = r2

h2 + k2 + 65 – 16h – 2k = r2 ………….(iv)

Subtract (ii) from (iii), we get

h – 7k – 26 = 0 ……(v)

Subtract (ii) from (iv), we get

7h + k – 32 = 0 ……(vi)

Solving (v) and (vi)

h = 5 and k = – 3

Equation (iv) ⇒ r = 25

[using h = 5 and k = – 3]

Therefore,

Centre (5, – 3)

Radius = 25

Check for (9, – 6):

To check if (9, – 6) lies on the circle,

(9 – 5)2 + ( – 6 + 3)2 = 52

25 = 25

Which is true.

Hence, all the points are lie on circle.


R S Aggarwal Solutions For Class 11 Maths Chapter 21 Exercises:

Get detailed solutions for all the questions listed under below exercises:

Exercise 21A Solutions

Exercise 21B Solutions

R S Aggarwal Solutions for Chapter 21 Circles

R S Aggarwal Chapter 21 – Circles, contains the answers to all the questions listed in this chapter. This chapter is a set of two exercises where students will learn about various concepts related to circle and its equation.

Let us have a look at some of the topics that are being covered here:

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