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} = r^{2 }

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

x^{2} + y^{2} = r^{2}

3. The equation of a circle with A(x_{1} , y_{1} ) and B(x_{2} , y_{2}) as the end points of a diameter is given by (x – x_{1})(x – x_{2}) + (y – y_{1})(y – y_{2}) = 0

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

## Download PDF of R S Aggarwal Solutions for Class 11 Maths Chapter 21 Circles

### 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} = r^{2} …..(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} = 5^{2}

or (x – 2)^{2 }+ (y – 4)^{2} = 25

or x^{2} + y^{2} – 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} = r^{2} …..(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} = 6^{2}

or (x + 3)^{2} + (y + 2)^{2} = 36

or x^{2} + y^{2} + 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} = r^{2} …..(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 x^{2} + y^{2} – 2ax – 2ay + (2a^{2} – 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} = r^{2} …..(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} = a^{2}

or x^{2} + y^{2} – 2a cos ∝ x – 2a sin ∝ y + a^{2} cos^{2} ∝ + a^{2} sin^{2} ∝ = a^{2}

or x^{2} + y^{2} – 2a cos ∝ x – 2a sin ∝ y = 0

^{2}∝ + a

^{2}sin

^{2}∝ = 1]

Which is the required equation.

**Question 5: Find the equation of a circle with centre (-a, -b) and radius √(a ^{2} – b^{2}).**

**Solution: **

The general form of the equation of a circle is:

(x – h)^{2} + (y – k)^{2} = r^{2} …..(1)

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

Here, r = √(a^{2} – b^{2}), h = -a and k = -b

Equation (1)⇒

(x + a )^{2} + (y + b)^{2} = (√(a^{2} – b^{2}))^{2}

or (x + a )^{2} + (y + b)^{2 }= a^{2 } – b^{2}

or x^{2} + y^{2} + 2a x + 2ay + a^{2} + b^{2} = a^{2 } – b^{2}

or x^{2} + y^{2} + 2a x + 2ay + 2 b^{2} = 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} = r^{2} …..(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 x^{2} + y^{2 }= 16

or x^{2} + y^{2 }– 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) **

**(iii) (x + 5) ^{2} + (y – 3)^{ 2} = 20**

**(iv) x ^{2} + (y – 1)^{ 2} = 2**

**Solution: **

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

(x – h)^{2} + (y – k)^{ 2} = r^{2}

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, r^{2} = 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} = r^{2}

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

r = radius of the circle.

Given equation is

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

h = 1/2 , k = – 1/3, r^{2} = 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} = r^{2}

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, r^{2} = 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} = r^{2}

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

r = radius of the circle.

Given equation is

x^{2} + (y – 1)^{ 2} = 2

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

h = 0 , k = 1, r^{2} = 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} = r^{2 } …..(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} = r^{2 }

1 + 49 = r^{2}

Or r^{2} = 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 x ^{2} + y^{2} – 4x + 6y – 5 = 0 represents a circle. Find its centre and radius.**

**Solution: **

Given equation is x^{2} + y^{2} – 4x + 6y – 5 = 0

The general equation of a circle is as follows:

x^{2} + y^{2} + 2gx + 2fy + c = 0

Where g, f and c are constants

With

Centre: (-g, -f)

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)

**Question 2: Show that the equation x ^{2} + y^{2} + x – y = 0 represents a circle. Find its centre and radius.**

**Solution: **

Given equation is x^{2} + y^{2} + x – y = 0

The general equation of a circle is as follows:

x^{2} + y^{2} + 2gx + 2fy + c = 0

Where g, f and c are constants

With

Centre: (-g, -f)

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)

**Question 3: Show that the equation 3x ^{2} + 3y^{2} + 6x – 4y – 1 = 0 represents a circle. Find its centre and radius.**

**Solution: **

Given equation is 3x^{2} + 3y^{2} + 6x – 4y – 1 = 0

Or x^{2} + y^{2} + 2x – 4/3 y – 1/3 = 0

The general equation of a circle is as follows:

x^{2} + y^{2} + 2gx + 2fy + c = 0

Where g, f and c are constants.

With

Centre: (-g, -f)

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)

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

**Solution: **

Given equation is x^{2} + y^{2} + 2x + 10y + 26 = 0

The general equation of a circle:

x^{2} + y^{2} + 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).

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

**Question 5: Show that the equation x ^{2} + y^{2} – 3x + 3y + 10 = 0 does not represent a circle.**

**Solution: **

Given equation is x^{2} + y^{2} – 3x + 3y + 10 = 0

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

Find radius of given equation:

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:

x^{2} + y^{2} + 2gx + 2fy + c = 0

Where g, f and c are constants.

With

Centre: (-g, -f)

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

The Circle equation is:

Let us apply Laplace Expansion to solve this problem:

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)

**(ii) **(1, 2), (3, – 4) and (5, – 6)

The Circle equation is:

Let us apply Laplace Expansion to solve this problem:

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)

**(iii) **(20, 3), (19, 8) and (2, – 9)

The Circle equation is:

Let us apply Laplace Expansion to solve this problem:

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)

**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} = r^{2} …(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

h^{2} + k^{2} + 4h – 6k + 13 = r^{2} ……….(ii)

h^{2} + k^{2} – 10h – 4k + 29 = r^{2} ………(iii)

h^{2} + k^{2} – 12h + 2k + 37 = r^{2} ………(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} = 5^{2}

(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, x^{2} + y^{2} + 4x + 6y + 11 = 0

The concentric circle will have the equation

x^{2} + y^{2} + 4x + 6y + d = 0 ….(1)

As it passes through P(5, 4),

Put x = 5 and y = 4

5^{2} + 4^{2} + 20 + 24 + d= 0

25 + 16 + 20 + 24 + d = 0

d = – 85

Equation (1) ⇒ x^{2} + y^{2} + 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} = r^{2} …(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) ⇒ h^{2} + k^{2} + 1 – 2h = r^{2} ……(ii)

Putting (2, – 7) ⇒ h^{2} + k^{2} + 53 – 4h + 14k = r^{2} …….(iii)

Putting (8, 1) ⇒ (8 – h)^{2} + (1 – k)^{ 2} = r^{2}

h^{2} + k^{2} + 65 – 16h – 2k = r^{2 } ………….(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} = 5^{2}

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:

## 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