## RD Sharma Solutions Class 11 Maths Chapter 24 – Avail Free PDF Updated for 2023 – 24

**RD Sharma Solutions for Class 11 Maths Chapter 24 – The Circle** are provided here for students to learn the concepts comprehensively and score good marks in the board exams. In this section, we shall study the circle and also find the equation of any circle whose centre and radius are given. Subject experts at BYJU’S have formulated the **RD Sharma Class 11 Solutions for Maths** in the most lucid and easy manner based on the latest CBSE syllabus and guidelines provided by the CBSE board. Students who find difficulty in understanding the concepts can refer to RD Sharma Class 11 Maths Solutions, which are the best reference materials.

Chapter 24 – The Circle contains three exercises, and RD Sharma Solutions provide precise answers with illustrations to the questions present in each exercise. The **RD Sharma Solutions** are prepared using shortcut techniques to help students grasp the concepts faster and make learning fun and interesting. The solutions to this chapter are available in PDF format, which can be downloaded easily from the links provided below. Now, let us have a look at the concepts discussed in this chapter.

- Standard equation of a circle
- Some particular cases
- General equations of a circle
- Diameter form of a circle

## RD Sharma Solutions for Class 11 Maths Chapter 24 – The Circle

### Access answers to RD Sharma Solutions for Class 11 Maths Chapter 24 – The Circle

EXERCISE 24.1 PAGE NO: 24.21

**1. Find the equation of the circle with:
(i) Centre (-2, 3) and radius 4.**

**(ii) Centre (a, b) and radius.**

**(iii) Centre (0, – 1) and radius 1.**

**(iv) Centre (a cos α, a sin α) and radius a.**

**(v) Centre (a, a) and radius √2 a.**

**Solution:**

**(i) **Centre (-2, 3) and radius 4.

Given:

The radius is 4, and the centre (-2, 3)

By using the formula,

The equation of the circle with centre (p, q) and radius ‘r’ is (x – p)^{2} + (y – q)^{2} = r^{2}

Where, p = -2, q = 3, r = 4

Now, by substituting the values in the above equation, we get

(x – p)^{2} + (y – q)^{2} = r^{2}

(x – (-2))^{2} + (y – 3)^{2} = 4^{2}

(x + 2)^{2} + (y – 3)^{2} = 16

x^{2} + 4x + 4 + y^{2} – 6y + 9 = 16

x^{2} + y^{2} + 4x – 6y – 3 = 0

*∴ The equation of the circle is x ^{2} + y^{2} + 4x – 6y – 3 = 0*

**(ii) **Centre (a, b) and radius

.

Given:

The radius is

and the centre (a, b)

By using the formula,

The equation of the circle with centre (p, q) and radius ‘r’ is (x – p)^{2} + (y – q)^{2} = r^{2}

Where, p = a, q = b, r =

Now, by substituting the values in the above equation, we get

(x – p)^{2} + (y – q)^{2} = r^{2}

(x – a)^{2} + (y – b)^{2} =

x^{2} – 2ax + a^{2} + y^{2} – 2by + b^{2} = a^{2} + b^{2}

x^{2} + y^{2} – 2ax – 2by = 0

*∴ The equation of the circle is x ^{2} + y^{2} – 2ax – 2by = 0*

**(iii) **Centre (0, -1) and radius 1.

Given:

The radius is 1, and the centre (0, -1)

By using the formula,

The equation of the circle with centre (p, q) and radius ‘r’ is (x – p)^{2} + (y – q)^{2} = r^{2}

Where, p = 0, q = -1, r = 1

Now, by substituting the values in the above equation, we get

(x – p)^{2} + (y – q)^{2} = r^{2}

(x – 0)^{2} + (y – (-1))^{2} = 1^{2}

(x – 0)^{2} + (y + 1)^{2} = 1

x^{2} + y^{2} + 2y + 1 = 1

x^{2} + y^{2} + 2y = 0

*∴ The equation of the circle is x ^{2} + y^{2} + 2y = 0.*

**(iv) **Centre (a cos α, a sin α) and radius a.

Given:

The radius is ‘a’, and the centre (a cos α, a sin α)

By using the formula,

The equation of the circle with centre (p, q) and radius ‘r’ is (x – p)^{2} + (y – q)^{2} = r^{2}

Where, p = a cos α, q = a sin α, r = a

Now, by substituting the values in the above equation, we get

(x – p)^{2} + (y – q)^{2} = r^{2}

(x – a cosα)^{2} + (y – a sinα)^{2} = a^{2}

x^{2} – (2acosα)x + a^{2}cos^{2}α + y^{2} – (2asinα)y + a^{2}sin^{2}α = a^{2}

We know that sin^{2}θ + cos^{2}θ = 1

So,

x^{2} – (2acosα)x + y^{2} – 2asinαy + a^{2} = a^{2}

x^{2} + y^{2} – (2acosα)x – (2asinα)y = 0

*∴ The equation of the circle is x ^{2} + y^{2} – (2acosα) x – (2asinα) y = 0.*

**(v) **Centre (a, a) and radius √2 a.

Given:

The radius is √2 a, and the centre (a, a)

By using the formula,

The equation of the circle with centre (p, q) and radius ‘r’ is (x – p)^{2} + (y – q)^{2} = r^{2}

Where, p = a, q = a, r = √2 a

Now, by substituting the values in the above equation, we get

(x – p)^{2} + (y – q)^{2} = r^{2}

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

x^{2} – 2ax + a^{2} + y^{2} – 2ay + a^{2} = 2a^{2}

x^{2} + y^{2} – 2ax – 2ay = 0

*∴ The equation of the circle is x ^{2} + y^{2} – 2ax – 2ay = 0.*

**2. Find the centre and radius of each of the following circles:
(i) (x – 1) ^{2} + y^{2} = 4**

**(ii) (x + 5) ^{2} + (y + 1)^{2} = 9**

**(iii) x ^{2} + y^{2} – 4x + 6y = 5**

**(iv) x ^{2} + y^{2} – x + 2y – 3 = 0**

**Solution:**

**(i) **(x – 1)^{2} + y^{2} = 4

Given:

The equation (x – 1)^{2} + y^{2} = 4

We need to find the centre and the radius.

By using the standard equation formula,

(x – a)^{2} + (y – b)^{2} = r^{2} …. (1)

Now, let us convert the given circle’s equation into the standard form.

(x – 1)^{2} + y^{2} = 4

(x – 1)^{2} + (y – 0)^{2} = 2^{2} ….. (2)

By comparing equation (2) with (1), we get

Centre = (1, 0) and radius = 2

*∴ The centre of the circle is (1, 0), and the radius is 2.*

**(ii) **(x + 5)^{2} + (y + 1)^{2} = 9

Given:

The equation (x + 5)^{2} + (y + 1)^{2} = 9

We need to find the centre and the radius.

By using the standard equation formula,

(x – a)^{2} + (y – b)^{2} = r^{2} …. (1)

Now let us convert the given circle’s equation into the standard form.

(x + 5)^{2} + (y + 1)^{2} = 9

(x – (-5))^{2} + (y – ( – 1))^{2} = 3^{2} …. (2)

By comparing equation (2) with (1), we get

Centre = (-5, -1) and radius = 3

*∴ The centre of the circle is (-5, -1), and the radius is 3.*

**(iii) **x^{2} + y^{2} – 4x + 6y = 5

Given:

The equation x^{2} + y^{2} – 4x + 6y = 5

We need to find the centre and the radius.

By using the standard equation formula,

(x – a)^{2} + (y – b)^{2} = r^{2} …. (1)

Now let us convert the given circle’s equation into the standard form.

x^{2} + y^{2} – 4x + 6y = 5

(x^{2} – 4x + 4) + (y^{2} + 6y + 9) = 5 + 4 + 9

(x – 2)^{2} + (y + 3)^{2} = 18

(x – 2)^{2} + (y – (-3))^{2} = (3√2)^{2} … (2)

By comparing equation (2) with (1), we get

Centre = (2, -3) and radius = 3√2

*∴ The centre of the circle is (2, -3), and the radius is 3√2.*

**(iv) **x^{2} + y^{2} – x + 2y – 3 = 0

Given:

The equation x^{2} + y^{2} – x + 2y – 3 = 0

We need to find the centre and the radius.

By using the standard equation formula,

(x – a)^{2} + (y – b)^{2} = r^{2} …. (1)

Now let us convert the given circle’s equation into the standard form.

x^{2} + y^{2} – x + 2y – 3 = 0

(x^{2} – x + ¼) + (y^{2} + 2y + 1) – 3 – ¼ – 1 = 0

(x – ½)^{2} + (y + 1)^{2} = 17/4 …. (2)

By comparing equation (2) with (1), we get

Centre = (½, – 1) and radius = √17/2

*∴ The centre of the circle is (½, -1), and the radius is √17/2.*

**3. Find the equation of the circle whose centre is (1, 2) and which passes through the point (4, 6).**

**Solution:**

Given:

Centre is (1, 2) and passes through the point (4, 6).

Where, p = 1, q = 2

We need to find the equation of the circle.

By using the formula,

(x – p)^{2} + (y – q)^{2} = r^{2}

(x – 1)^{2} + (y – 2)^{2} = r^{2}

It passes through the point (4, 6)

(4 – 1)^{2} + (6 – 2)^{2} = r^{2}

3^{2} + 4^{2} = r^{2}

9 + 16 = r^{2}

25 = r^{2}

r = √25

= 5

So r = 5 units

We know that the equation of the circle with centre (p, q) and having radius ‘r’ is given by: (x – p)^{2} + (y – q)^{2} = r^{2}

By substituting the values in the above equation, we get

(x – 1)^{2} + (y – 2)^{2} = 5^{2}

x^{2} – 2x + 1 + y^{2} – 4y + 4 = 25

x^{2} + y^{2} – 2x – 4y – 20 = 0.

*∴ The equation of the circle is x ^{2} + y^{2} – 2x – 4y – 20 = 0.*

**4. Find the equation of the circle passing through the point of intersection of the lines x + 3y = 0 and 2x – 7y = 0 and whose centre is the point of intersection of the lines x + y + 1 = 0 and x – 2y + 4 = 0.**

**Solution:**

Let us find the points of intersection of the lines.

On solving the lines x + 3y = 0 and 2x – 7y = 0, we get the point of intersection to be (0, 0)

On solving the lines x + y + 1 and x – 2y + 4 = 0, we get the point of intersection to be (-2, 1)

We have a circle with a centre (-2, 1) and passing through the point (0, 0).

We know that the radius of the circle is the distance between the centre and any point on the radius. So, we find the radius of the circle.

So, the equation is (x – p)^{2} + (y – q)^{2} = r^{2}

Where, p = -2, q = 1

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

Equation (1) passes through (0, 0)

So, (0 + 2)^{2} + (0 – 1)^{2} = r^{2}

4 + 1 = a^{2}

5 = r^{2}

r = √5

We know that the equation of the circle with centre (p, q) and having radius ‘r’ is given by: (x – p)^{2} + (y – q)^{2} = r^{2}

By substituting the values in the above equation, we get

(x – (-2))^{2} + (y – 1)^{2} = (√5)^{2}

(x + 2)^{2} + (y – 1)^{2} = 5

x^{2} + 4x + 4 + y^{2} – 2y + 1 = 5

x^{2} + y^{2} + 4x – 2y = 0

*∴ The equation of the circle is x ^{2} + y^{2} + 4x – 2y = 0.*

**5. Find the equation of the circle whose centre lies on the positive direction of the y-axis at a distance 6 from the origin and whose radius is 4.**

**Solution: **

It is given that the centre lies on the positive y-axis at a distance of 6 from the origin, we get the centre (0, 6).

We have a circle with a centre (0, 6) and having radius 4.

We know that the equation of the circle with centre (p, q) and having radius ‘r’ is given by: (x – p)^{2} + (y – q)^{2} = r^{2}

Where, p = 0, q = 6, r = 4

Now, by substituting the values in the equation, we get

(x – 0)^{2} + (y – 6)^{2} = 4^{2}

x^{2} + y^{2} – 12y + 36 = 16

x^{2} + y^{2} – 12y + 20 = 0.

*∴ The equation of the circle is x ^{2} + y^{2} – 12y + 20 = 0.*

**6. If the equations of two diameters of a circle are 2x + y = 6 and 3x + 2y = 4 and the radius is 10, find the equation of the circle.**

**Solution:**

It is given that the circle has a radius of 10 and has diameters 2x + y = 6 and 3x + 2y = 4.

We know that the centre is the intersection point of the diameters.

On solving the diameters, we get the centre to be (8, -10).

We have a circle with a centre (8, -10) and having radius 10.

By using the formula,

^{2} + (y – q)^{2} = r^{2}

Where, p = 8, q = -10, r = 10

Now, by substituting the values in the equation, we get

(x – 8)^{2} + (y – (-10))^{2} = 10^{2}

(x – 8)^{2} + (y + 10)^{2} = 100

x^{2} – 16x + 64 + y^{2} + 20y + 100 = 100

x^{2} + y^{2} – 16x + 20y + 64 = 0.

*∴ The equation of the circle is x ^{2} + y^{2} – 16x + 20y + 64 = 0.*

**7.** **Find the equation of the circle
(i) which touches both axes at a distance of 6 units from the origin.**

**(ii) Which touches the x-axis at a distance of 5 from the origin and radius 6 units.**

**(iii) Which touches both the axes and passes through the point (2, 1).**

**(iv) Passing through the origin, radius 17 and ordinate of the centre is – 15.**

**Solution:**

**(i) **which touches both axes at a distance of 6 units from the origin.

A circle touches the axes at the points (±6, 0) and (0, ±6).

So, a circle has a centre (±6, ±6) and passes through the point (0, 6).

We know that the radius of the circle is the distance between the centre and any point on the radius. So, we find the radius of the circle.

So, the equation is (x – p)^{2} + (y – q)^{2} = r^{2}

Where, p = 6, q = 6

(x – 6)^{2} + (y – 6)^{2} = r^{2} …. (1)

Equation (1) passes through (0, 6)

So, (0 – 6)^{2} + (6 – 6)^{2} = r^{2}

36 + 0 = r^{2}

r = √36

= 6

^{2} + (y – q)^{2} = r^{2}

Now by substituting the values in the equation, we get

(x ± 6)^{2} + (y ± 6)^{2} = (6)^{2}

x^{2 }± 12x + 36 + y^{2 }± 12y + 36 = 36

x^{2} + y^{2 }± 12x ± 12y + 36 = 0

*∴ The equation of the circle is x ^{2} + y^{2 }± 12x ± 12y + 36 = 0.*

**(ii) **Which touches the x -axis at a distance of 5 from the origin and radius 6 units.

A circle touches the x -axis at the points (±5, 0).

Let us assume the centre of the circle is (±5, a).

We have a circle with a centre (5, a) and passing through the point (5, 0) and having radius 6.

We know that the radius of the circle is the distance between the centre and any point on the radius. So, we find the radius of the circle.

So, the equation is (x – p)^{2} + (y – q)^{2} = r^{2}

Where, p = 5, q = a

(x – 5)^{2} + (y – a)^{2} = r^{2} …. (1)

Equation (1) passes through (5, 0)

So, (5 – 5)^{2} + (0 – 6)^{2} = r^{2}

0 + 36 = r^{2}

r = √36

= 6

We have got the centre at (±5, ±6) and having radius 6 units.

^{2} + (y – q)^{2} = r^{2}

Now by substituting the values in the equation, we get

(x ± 5)^{2} + (y ± 6)^{2} = (6)^{2}

x^{2} ± 10x + 25 + y^{2} ± 12y + 36 = 36

x^{2} + y^{2} ± 10x ± 12y + 25 = 0.

*∴ The equation of the circle is x ^{2} + y^{2} ± 10x ± 12y + 25 = 0.*

**(iii) **Which touches both the axes and passes through the point (2, 1).

Let us assume the circle touches the x-axis at the point (a, 0) and the y-axis at the point (0, a).

Then the centre of the circle is (a, a), and the radius is a.

Its equation will be (x – p)^{2} + (y – q)^{2} = r^{2}

By substituting the values, we get

(x – a)^{2} + (y – a)^{2} = a^{2} … (1)

So now, equation (1) passes through P (2, 1)

By substituting the values, we get

(2 – a)^{2} + (1 – a)^{2} = a^{2}

4 – 4a + a^{2} + 1 – 2a + a^{2} = a^{2}

5 – 6a + a^{2} = 0

(a – 5) (a – 1) = 0

So, a = 5 or 1

Case (i)

We have got the centre at (5, 5) and having radius 5 units.

^{2} + (y – q)^{2} = r^{2}

Now by substituting the values in the equation, we get

(x – 5)^{2} + (y – 5)^{2} = 5^{2}

x^{2} – 10x + 25 + y^{2} – 10y + 25 = 25

x^{2} + y^{2} – 10x – 10y + 25 = 0.

*∴ The equation of the circle is x ^{2} + y^{2} – 10x – 10y + 25 = 0.*

Case (ii)

We have got the centre at (1, 1) and having a radius 1 unit.

^{2} + (y – q)^{2} = r^{2}

Now by substituting the values in the equation, we get

(x – 1)^{2} + (y – 1)^{2} = 1^{2}

x^{2} – 2x + 1 + y^{2} – 2y + 1 = 1

x^{2} + y^{2} – 2x – 2y + 1 = 0

*∴ The equation of the circle is x ^{2} + y^{2} – 2x – 2y + 1 = 0.*

**(iv) **Passing through the origin, radius 17 and ordinate of the centre is – 15.

Let us assume the abscissa as ‘a’.

We have a circle with a centre (a, – 15) and passing through the point (0, 0) and having radius 17.

By using the distance formula,

17^{2} = a^{2} + (-15)^{2}

289 = a^{2} + 225

a^{2} = 64

|a| = √64

|a| = 8

a = ±8 …. (1)

We have got the centre at (±8, – 15) and having radius 17 units.

^{2} + (y – q)^{2} = r^{2}

Now, by substituting the values in the equation, we get

(x ± 8)^{2} + (y – 15)^{2} = 17^{2}

x^{2 }± 16x + 64 + y^{2} – 30y + 225 = 289

x^{2} + y^{2 }± 16x – 30y = 0.

*∴ The equation of the circle is x ^{2} + y^{2 }± 16x – 30y = 0.*

**8. Find the equation of the circle which has its centre at the point (3, 4) and touches the straight line 5x + 12y – 1 = 0.**

**Solution:**

It is given that we need to find the equation of the circle with centre (3, 4) and touches the straight line 5x + 12y – 1 = 0.

We have a circle with a centre (3, 4) and having a radius 62/13.

^{2} + (y – q)^{2} = r^{2}

Now, by substituting the values in the equation, we get

(x – 3)^{2} + (y – 4)^{2} = (62/13)^{2}

x^{2} – 6x + 9 + y^{2} – 8y + 16 = 3844/169

169x^{2} + 169y^{2} – 1014x – 1352y + 4225 = 3844

169x^{2} + 169y^{2} – 1014x – 1352y + 381 = 0

*∴ The equation of the circle is 169x ^{2} + 169y^{2} – 1014x – 1352y + 381 = 0.*

**9. Find the equation of the circle which touches the axes and whose centre lies on x – 2y = 3.**

**Solution:**

Let us assume the circle touches the axes at (a, 0) and (0, a), and we get the radius to be |a|.

We get the centre of the circle as (a, a). This point lies on the line x – 2y = 3

a – 2(a) = 3

-a = 3

a = – 3

Centre = (a, a) = (-3, -3) and radius of the circle(r) = |-3| = 3

We have the circle with a centre (-3, -3) and having radius 3.

^{2} + (y – q)^{2} = r^{2}

Now, by substituting the values in the equation, we get

(x – (-3))^{2} + (y – (-3))^{2} = 3^{2}

(x + 3)^{2} + (y + 3)^{2} = 9

x^{2} + 6x + 9 + y^{2} + 6y + 9 = 9

x^{2} + y^{2} + 6x + 6y + 9 = 0

*∴ The equation of the circle is x ^{2} + y^{2} + 6x + 6y + 9 = 0.*

**10. A circle whose centre is the point of intersection of the lines 2x – 3y + 4 = 0 and 3x + 4y – 5 = 0 passes through the origin. Find its equation.**

**Solution:**

It is given that the circle has the centre at the intersection point of the lines 2x – 3y + 4 = 0 and 3x + 4y – 5 = 0 and passes through the origin

^{2} + (y – q)^{2} = r^{2}

Now, by substituting the values in the equation, we get

EXERCISE 24.2 PAGE NO: 24.31

**1. Find the coordinates of the centre radius of each of the following circles:
(i) x ^{2} + y^{2} + 6x – 8y – 24 = 0**

**(ii) 2x ^{2} + 2y^{2} – 3x + 5y = 7**

**(iv) x ^{2} + y^{2} – ax – by = 0**

**Solution:**

**(i) **x^{2} + y^{2} + 6x – 8y – 24 = 0

Given:

The equation of the circle is x^{2} + y^{2} + 6x – 8y – 24 = 0 …… (1)

We know that for a circle x^{2} + y^{2} + 2ax + 2by + c = 0 …… (2)

Centre = (-a, -b)

So by comparing equations (1) and (2),

Centre = (-6/2, -(-8)/2)

= (-3, 4)

Radius = **√**(a^{2} + b^{2} – c)

= **√**(3^{2} + 4^{2} – (-24))

= **√**(9 + 16 + 24)

= **√**(49)

= 7

*∴ The centre of the circle is (-3, 4), and the radius is 7.*

**(ii) **2x^{2} + 2y^{2} – 3x + 5y = 7

Given:

The equation of the circle is 2x^{2} + 2y^{2} – 3x + 5y = 7 (divide by 2 we get)

x^{2} + y^{2} – 3x/2 + 5y/2 = 7/2

Now, by comparing with the equation x^{2} + y^{2} + 2ax + 2by + c = 0

Centre = (-a, -b)

Given:

The equation of the circle is

(Multiply by 2 we get)

x^{2} + y^{2} + 2x cos θ + 2y sin θ – 8 = 0

By comparing with the equation x^{2} + y^{2} + 2ax + 2by + c = 0

Centre = (-a, -b)

= [(-2cos θ)/2 , (-2sin θ)/2]

= (-cos θ, -sin θ)

Radius = **√**(a^{2} + b^{2} – c)

= √[(-cos θ)^{2} + (sin θ)^{2} – (-8)]

= √[cos^{2} θ + sin^{2} θ + 8]

= √[1 + 8]

= √[9]

= 3

*∴ The centre and radius of the circle are (-cos θ, -sin θ) and 3.*

**(iv) **x^{2} + y^{2} – ax – by = 0

Given:

Equation of the circle is x^{2} + y^{2} – ax – by = 0

By comparing with the equation x^{2} + y^{2} + 2ax + 2by + c = 0

Centre = (-a, -b)

= (-(-a)/2, -(-b)/2)

= (a/2, b/2)

Radius = **√**(a^{2} + b^{2} – c)

= √[(a/2)^{2} + (b/2)^{2}]

= **√**[(a^{2}/4 + b^{2}/4)]

= **√**[(a^{2} + b^{2})/4]

= [**√**(a^{2} + b^{2})]/2

*∴ The centre and radius of the circle is (a/2, b/2) and [ √(a^{2} + b^{2})]/2*

**2. Find the equation of the circle passing through the points :
(i) (5, 7), (8, 1) and (1, 3)**

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

**(iii) (5, -8), (-2, 9) and (2, 1)**

**(iv) (0, 0), (-2, 1) and (-3, 2)**

**Solution:**

**(i) (5, 7), (8, 1) and (1, 3)**

By using the standard form of the equation of the circle:

x^{2} + y^{2} + 2ax + 2by + c = 0 ….. (1)

Firstly let us find the values of a, b and c

Substitute the given point (5, 7) in equation (1), and we get

5^{2} + 7^{2} + 2a (5) + 2b (7) + c = 0

25 + 49 + 10a + 14b + c = 0

10a + 14b + c + 74 = 0….. (2)

By substituting the given point (8, 1) in equation (1), we get

8^{2} + 1^{2} + 2a (8) + 2b (1) + c = 0

64 + 1 + 16a + 2b + c = 0

16a + 2b + c + 65 = 0….. (3)

Substituting the point (1, 3) in equation (1), we get

1^{2} + 3^{2} + 2a (1) + 2b (3) + c = 0

1 + 9 + 2a + 6b + c = 0

2a + 6b + c + 10 = 0….. (4)

Now by simplifying the equations (2), (3), (4), we get the values

a = -29/6, b = -19/6, c = 56/3

Substituting the values of a, b, c in equation (1), we get

x^{2} + y^{2} + 2 (-29/6)x + 2 (-19/6) + 56/3 = 0

x^{2} + y^{2} – 29x/3 – 19y/3 + 56/3 = 0

3x^{2} + 3y^{2} – 29x – 19y + 56 = 0

*∴ The equation of the circle is 3x ^{2} + 3y^{2} – 29x – 19y + 56 = 0*

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

By using the standard form of the equation of the circle:

x^{2} + y^{2} + 2ax + 2by + c = 0 ….. (1)

Substitute the points (1, 2) in equation (1), and we get

1^{2} + 2^{2} + 2a (1) + 2b (2) + c = 0

1 + 4 + 2a + 4b + c = 0

2a + 4b + c + 5 = 0….. (2)

Substitute the points (3, -4) in equation (1), and we get

3^{2} + (- 4)^{2} + 2a (3) + 2b (- 4) + c = 0

9 + 16 + 6a – 8b + c = 0

6a – 8b + c + 25 = 0….. (3)

Substitute the points (5, -6) in equation (1), and we get

5^{2} + (- 6)^{2} + 2a (5) + 2b (- 6) + c = 0

25 + 36 + 10a – 12b + c = 0

10a – 12b + c + 61 = 0….. (4)

Now by simplifying the equations (2), (3), (4), we get

a = – 11, b = – 2, c = 25

Substitute the values of a, b and c in equation (1), and we get

x^{2} + y^{2} + 2(- 11)x + 2(- 2) + 25 = 0

x^{2} + y^{2} – 22x – 4y + 25 = 0

*∴ The equation of the circle is x ^{2} + y^{2} – 22x – 4y + 25 = 0*

**(iii) (5, -8), (-2, 9) and (2, 1)**

By using the standard form of the equation of the circle,

x^{2} + y^{2} + 2ax + 2by + c = 0 ….. (1)

Substitute the point (5, -8) in equation (1), and we get

5^{2} + (- 8)^{2} + 2a(5) + 2b(- 8) + c = 0

25 + 64 + 10a – 16b + c = 0

10a – 16b + c + 89 = 0….. (2)

Substitute the points (-2, 9) in equation (1), and we get

(- 2)^{2} + 9^{2} + 2a(- 2) + 2b(9) + c = 0

4 + 81 – 4a + 18b + c = 0

-4a + 18b + c + 85 = 0….. (3)

Substitute the points (2, 1) in equation (1), and we get

2^{2} + 1^{2} + 2a(2) + 2b(1) + c = 0

4 + 1 + 4a + 2b + c = 0

4a + 2b + c + 5 = 0….. (4)

By simplifying equations (2), (3), (4), we get

a = 58, b = 24, c = – 285.

Now, by substituting the values of a, b, c in equation (1), we get

x^{2} + y^{2} + 2(58)x + 2(24) – 285 = 0

x^{2} + y^{2} + 116x + 48y – 285 = 0

*∴ The equation of the circle is x ^{2} + y^{2} + 116x + 48y – 285 = 0*

**(iv) (0, 0), (-2, 1) and (-3, 2)**

By using the standard form of the equation of the circle:

x^{2} + y^{2} + 2ax + 2by + c = 0 ….. (1)

Substitute the points (0, 0) in equation (1), and we get

0^{2} + 0^{2} + 2a(0) + 2b(0) + c = 0

0 + 0 + 0a + 0b + c = 0

c = 0….. (2)

Substitute the points (-2, 1) in equation (1), and we get

(- 2)^{2} + 1^{2} + 2a(- 2) + 2b(1) + c = 0

4 + 1 – 4a + 2b + c = 0

-4a + 2b + c + 5 = 0….. (3)

Substitute the points (-3, 2) in equation (1), and we get

(- 3)^{2} + 2^{2} + 2a(- 3) + 2b(2) + c = 0

9 + 4 – 6a + 4b + c = 0

-6a + 4b + c + 13 = 0….. (4)

By simplifying the equations (2), (3), (4), we get

a = -3/2, b = -11/2, c = 0

Now, by substituting the values of a, b, c in equation (1), we get

x^{2} + y^{2} + 2(-3/2)x + 2(-11/2)y + 0 = 0

x^{2} + y^{2} – 3x – 11y = 0

*∴ The equation of the circle is x ^{2} + y^{2} – 3x – 11y = 0*

**3. Find the equation of the circle which passes through (3, – 2), (- 2, 0) and has its centre on the line 2x – y = 3.**

**Solution:**

Given:

The line 2x – y = 3 … (1)

The points (3, -2), (-2, 0)

By using the standard form of the equation of the circle,

x^{2} + y^{2} + 2ax + 2by + c = 0 ….. (2)

Let us substitute the centre (-a, -b) in equation (1) we get,

2(- a) – (- b) = 3

-2a + b = 3

2a – b + 3 = 0…… (3)

Now Substitute the given points (3, -2) in equation (2), and we get

3^{2} + (- 2)^{2} + 2a(3) + 2b(- 2) + c = 0

9 + 4 + 6a – 4b + c = 0

6a – 4b + c + 13 = 0….. (4)

Substitute the points (-2, 0) in equation (2), and we get

(- 2)^{2} + 0^{2} + 2a(- 2) + 2b(0) + c = 0

4 + 0 – 4a + c = 0

4a – c – 4 = 0….. (5)

By simplifying the equations (3), (4) and (5), we get,

a = 3/2, b = 6, c = 2

Again by substituting the values of a, b, c in (2), we get

x^{2} + y^{2} + 2 (3/2)x + 2 (6)y + 2 = 0

x^{2} + y^{2} + 3x + 12y + 2 = 0

*∴ The equation of the circle is x ^{2} + y^{2} + 3x + 12y + 2 = 0.*

**4. Find the equation of the circle which passes through the points (3, 7), (5, 5) and has its centre on line x – 4y = 1.**

**Solution:**

Given:

Points (3, 7), (5, 5)

The line x – 4y = 1…. (1)

By using the standard form of the equation of the circle,

x^{2} + y^{2} + 2ax + 2by + c = 0 ….. (2)

Let us substitute the centre (-a, -b) in equation (1) we get,

(- a) – 4(- b) = 1

-a + 4b = 1

a – 4b + 1 = 0…… (3)

Substitute the points (3, 7) in equation (2), and we get

3^{2} + 7^{2} + 2a(3) + 2b(7) + c = 0

9 + 49 + 6a + 14b + c = 0

6a + 14b + c + 58 = 0….. (4)

Substitute the points (5, 5) in equation (2), and we get

5^{2} + 5^{2} + 2a(5) + 2b(5) + c = 0

25 + 25 + 10a + 10b + c = 0

10a + 10b + c + 50 = 0….. (5)

By simplifying equations (3), (4) and (5), we get,

a = 3, b = 1, c = – 90

Now, by substituting the values of a, b, c in equation (2), we get

x^{2} + y^{2} + 2(3)x + 2(1)y – 90 = 0

x^{2} + y^{2} + 6x + 2y – 90 = 0

*∴ The equation of the circle is x ^{2} + y^{2} + 6x + 2y – 90 = 0.*

**5. Show that the points (3, -2), (1, 0), (-1, -2) and (1, -4) are concyclic.**

**Solution:**

Given:

The points (3, -2), (1, 0), (-1, -2) and (1, -4)

Let us assume the circle passes through points A, B, C.

So by using the standard form of the equation of the circle,

x^{2} + y^{2} + 2ax + 2by + c = 0….. (1)

Substitute the points A (3, – 2) in equation (1), and we get,

3^{2} + (-2)^{2} + 2a(3) + 2b(-2) + c = 0

9 + 4 + 6a – 4b + c = 0

6a – 4b + c + 13 = 0….. (2)

Substitute the points B (1, 0) in equation (1), and we get,

1^{2} + 0^{2} + 2a(1) + 2b(0) + c = 0

1 + 2a + c = 0 ……- (3)

Substitute the points C (-1, -2) in equation (1), and we get,

(- 1)^{2} + (- 2)^{2} + 2a(- 1) + 2b(- 2) + c = 0

1 + 4 – 2a – 4b + c = 0

5 – 2a – 4b + c = 0

2a + 4b – c – 5 = 0….. (4)

Upon simplifying the equations (2), (3) and (4), we get,

a = – 1, b = 2 and c = 1

Substituting the values of a, b, c in equation (1), we get

x^{2} + y^{2} + 2(- 1)x + 2(2)y + 1 = 0

x^{2} + y^{2} – 2x + 4y + 1 = 0 ….. (5)

Now by substituting the point D (1, -4) in equation (5), we get,

1^{2} + (- 4)^{2} – 2(1) + 4(- 4) + 1

1 + 16 – 2 – 16 + 1

0

*∴ The points (3, -2), (1, 0), (-1, -2), (1, -4) are concyclic.*

**6. Show that the points (5, 5), (6, 4), (- 2, 4) and (7, 1) all lie on a circle, and find its equation, centre, and radius.**

**Solution:**

Given:

The points (5, 5), (6, 4), (- 2, 4) and (7, 1) all lie on a circle.

Let us assume the circle passes through points A, B, C.

So by using the standard form of the equation of the circle,

x^{2} + y^{2} + 2ax + 2by + c = 0….. (1)

Substituting A (5, 5) in (1), we get,

5^{2} + 5^{2} + 2a(5) + 2b(5) + c = 0

25 + 25 + 10a + 10b + c = 0

10a + 10b + c + 50 = 0….. (2)

Substitute the points B (6, 4) in equation (1), and we get,

6^{2} + 4^{2} + 2a(6) + 2b(4) + c = 0

36 + 16 + 12a + 8b + c = 0

12a + 8b + c + 52 = 0….. (3)

Substitute the point C (-2, 4) in equation (1), and we get,

(-2)^{2} + 4^{2} + 2a(-2) + 2b(4) + c = 0

4 + 16 – 4a + 8b + c = 0

20 – 4a + 8b + c = 0

4a – 8b – c – 20 = 0….. (4)

Upon simplifying equations (2), (3) and (4), we get,

a = – 2, b = – 1 and c = – 20

Now by substituting the values of a, b, c in equation (1), we get

x^{2} + y^{2} + 2(- 2)x + 2(- 1)y – 20 = 0

x^{2} + y^{2} – 4x – 2y – 20 = 0 ….. (5)

Substituting D (7, 1) in equation (5), we get,

7^{2} + 1^{2} – 4(7) – 2(1) – 20

49 + 1 – 28 – 2 – 20

0

∴ The points (3, -2), (1, 0), (-1, -2), (1, -4) lie on a circle.

Now, let us find the centre and the radius.

We know that for a circle x^{2} + y^{2} + 2ax + 2by + c = 0,

Centre = (-a, -b)

Radius = **√**(a^{2} + b^{2} – c)

Comparing equation (5) with equation (1), we get

Centre = [-(-4)/2, -(-2)/2)]

= (2, 1)

Radius = **√**(2^{2} + 1^{2} – (-20))

= **√**(25)

= 5

*∴ The centre and radius of the circle are (2, 1) and 5.*

**7. Find the equation of the circle which circumscribes the triangle formed by the lines:
(i) x + y + 3 = 0, x – y + 1 = 0 and x = 3**

**(ii) 2x + y – 3 = 0, x + y – 1 = 0 and 3x + 2y – 5 = 0**

**(iii) x + y = 2, 3x – 4y = 6 and x – y = 0**

**(iv) y = x + 2, 3y = 4x and 2y = 3x**

**Solution:**

**(i) x + y + 3 = 0, x – y + 1 = 0 and x = 3**

Given:

The lines x + y + 3 = 0

x – y + 1 = 0

x = 3

On solving these lines, we get the intersection points A (-2, -1), B (3, 4), C (3, -6)

So by using the standard form of the equation of the circle,

x^{2} + y^{2} + 2ax + 2by + c = 0….. (1)

Substitute the points (-2, -1) in equation (1), and we get

(- 2)^{2} + (- 1)^{2} + 2a(-2) + 2b(-1) + c = 0

4 + 1 – 4a – 2b + c = 0

5 – 4a – 2b + c = 0

4a + 2b – c – 5 = 0….. (2)

Substitute the points (3, 4) in equation (1), and we get

3^{2} + 4^{2} + 2a(3) + 2b(4) + c = 0

9 + 16 + 6a + 8b + c = 0

6a + 8b + c + 25 = 0….. (3)

Substitute the points (3, -6) in equation (1), and we get

3^{2} + (- 6)^{2} + 2a(3) + 2b(- 6) + c = 0

9 + 36 + 6a – 12b + c = 0

6a – 12b + c + 45 = 0….. (4)

Upon simplifying equations (2), (3), (4), we get

a = – 3, b = 1, c = -15.

Now, by substituting the values of a, b, c in equation (1), we get

x^{2} + y^{2} + 2(- 3)x + 2(1)y – 15 = 0

x^{2} + y^{2} – 6x + 2y – 15 = 0

*∴ The equation of the circle is x ^{2} + y^{2} – 6x + 2y – 15 = 0.*

**(ii) 2x + y – 3 = 0, x + y – 1 = 0 and 3x + 2y – 5 = 0**

Given:

The lines 2x + y – 3 = 0

x + y – 1 = 0

3x + 2y – 5 = 0

On solving these lines we get the intersection points A(2, – 1), B(3, – 2), C(1,1)

So by using the standard form of the equation of the circle:

x^{2} + y^{2} + 2ax + 2by + c = 0….. (1)

Substitute the points (2, -1) in equation (1), and we get

2^{2} + (- 1)^{2} + 2a(2) + 2b(- 1) + c = 0

4 + 1 + 4a – 2b + c = 0

4a – 2b + c + 5 = 0….. (2)

Substitute the points (3, -2) in equation (1), and we get

3^{2} + (- 2)^{2} + 2a(3) + 2b(- 2) + c = 0

9 + 4 + 6a – 4b + c = 0

6a – 4b + c + 13 = 0….. (3)

Substitute the points (1, 1) in equation (1), and we get

1^{2} + 1^{2} + 2a(1) + 2b(1) + c = 0

1 + 1 + 2a + 2b + c = 0

2a + 2b + c + 2 = 0….. (4)

Upon simplifying equations (2), (3), (4), we get

a = -13/2, b = -5/2, c = 16

Now, by substituting the values of a, b, c in equation (1), we get

x^{2} + y^{2} + 2 (-13/2)x + 2 (-5/2)y + 16 = 0

x^{2} + y^{2} – 13x – 5y + 16 = 0

*∴ The equation of the circle is x ^{2} + y^{2} – 13x – 5y + 16 = 0*

**(iii) x + y = 2, 3x – 4y = 6 and x – y = 0**

Given:

The lines x + y = 2

3x – 4y = 6

x – y = 0

On solving these lines we get the intersection points A(2,0), B(- 6, – 6), C(1,1)

So by using the standard form of the equation of the circle:

x^{2} + y^{2} + 2ax + 2by + c = 0….. (1)

Substitute the points (2, 0) in equation (1), and we get

2^{2} + 0^{2} + 2a(2) + 2b(0) + c = 0

4 + 4a + c = 0

4a + c + 4 = 0….. (2)

Substitute the point (-6, -6) in equation (1), and we get

(- 6)^{2} + (- 6)^{2} + 2a(- 6) + 2b(- 6) + c = 0

36 + 36 – 12a – 12b + c = 0

12a + 12b – c – 72 = 0….. (3)

Substitute the points (1, 1) in equation (1), and we get

1^{2} + 1^{2} + 2a(1) + 2b(1) + c = 0

1 + 1 + 2a + 2b + c = 0

2a + 2b + c + 2 = 0….. (4)

Upon simplifying equations (2), (3), (4), we get

a = 2, b = 3, c = – 12.

Substituting the values of a, b, c in equation (1), we get

x^{2} + y^{2} + 2(2)x + 2(3)y – 12 = 0

x^{2} + y^{2} + 4x + 6y – 12 = 0

*∴ The equation of the circle is x ^{2} + y^{2} + 4x + 6y – 12 = 0*

**(iv) y = x + 2, 3y = 4x and 2y = 3x**

Given:

The lines y = x + 2

3y = 4x

2y = 3x

On solving these lines we get the intersection points A(6,8), B(0,0), C(4,6)

So by using the standard form of the equation of the circle:

x^{2} + y^{2} + 2ax + 2by + c = 0….. (1)

Substitute the points (6, 8) in equation (1), and we get

6^{2} + 8^{2} + 2a(6) + 2b(8) + c = 0

36 + 64 + 12a + 16b + c = 0

12a + 16b + c + 100 = 0…… (2)

Substitute the points (0, 0) in equation (1), and we get

0^{2} + 0^{2} + 2a(0) + 2b(0) + c = 0

0 + 0 + 0a + 0b + c = 0

c = 0….. (3)

Substitute the points (4, 6) in equation (1), and we get

4^{2} + 6^{2} + 2a(4) + 2b(6) + c = 0

16 + 36 + 8a + 12b + c = 0

8a + 12b + c + 52 = 0….. (4)

Upon simplifying equations (2), (3), (4), we get

a = – 23, b = 11, c = 0

Now by substituting the values of a, b, c in equation (1), we get

x^{2} + y^{2} + 2(- 23)x + 2(11)y + 0 = 0

x^{2} + y^{2} – 46x + 22y = 0

*∴ The equation of the circle is x ^{2} + y^{2} – 46x + 22y = 0*

EXERCISE 24.3 PAGE NO: 24.37

**1. Find the equation of the circle, the endpoints whose diameter are (2, -3) and (-2, 4). Find its centre and radius.**

**Solution:**

Given:

The diameters are (2, -3) and (-2, 4).

By using the formula,

Centre = (-a, -b)

= [(2-2)/2, (-3+4)/2]

= (0, ½)

By using the distance formula,

So, r =

= **√**[(2-0)^{2} + (-3-½)^{2}]

= **√**[(2)^{2} + (-7/2)^{2}]

= **√**[4 + 49/4]

= **√**[65/4]

= [**√**65]/2

^{2} + (y – q)^{2} = r^{2}

Now, by substituting the values in the above equation, we get

(x – 0)^{2} + (y – ½)^{2} = [[**√**65]/2]^{2}

x^{2} + y^{2} – y + ¼ = 65/4

4x^{2} + 4y^{2} – 4y + 1 = 65

*∴The equation of the circle is 4x ^{2} + 4y^{2} – 4y – 64 = 0 or x^{2} + y^{2} – y – 16 = 0*

**2. Find the equation of the circle with the endpoints whose diameter are the centres of the circles x ^{2} + y^{2} + 6x – 14y – 1 = 0 and x^{2} + y^{2} – 4x + 10y – 2 = 0.**

**Solution:**

Given:

x^{2} + y^{2} + 6x – 14y – 1 = 0…. (1)

So the centre = [(-6/2), -(-14/2)]

= [-3, 7]

x^{2} + y^{2} – 4x + 10y – 2 = 0… (2)

So the centre = [-(-4/2), (-10/2)]

= [2, -5]

We know that the equation of the circle is given by,

(x – x_{1}) (x – x_{2}) + (y – y_{1}) (y – y_{2}) = 0

(x + 3) (x – 2) + (y – 7) (y + 5) = 0

Upon simplification, we get

x^{2} + 3x – 2x – 6 + y^{2} – 7y + 5y – 35 = 0

x^{2} + y^{2} + x – 2y – 41 = 0

*∴The equation of the circle is x ^{2} + y^{2} + x – 2y – 41 = 0*

**3. The sides of a squares are x = 6, x = 9, y = 3 and y = 6. Find the equation of a circle drawn on the diagonal of the square as its diameter.**

**Solution:**

Given:

The sides of the squares are x = 6, x = 9, y = 3 and y = 6.

Let us assume A, B, C, D be the vertices of the square. On solving the lines, we get the coordinates as A = (6, 3)

B = (9, 3)

C = (9, 6)

D = (6, 6)

We know that the equation of the circle with diagonal AC is given by

(x – x_{1}) (x – x_{2}) + (y – y_{1}) (y – y_{2}) = 0

(x – 6) (x – 9) + (4 – 3) (4 – 6) = 0

Upon simplifying, we get

x^{2} – 6x – 9x + 54 + y^{2} – 3y – 6y + 18 = 0

x^{2} + y^{2} – 15x – 9y + 72 = 0

We know that the equation of the circle with diagonal BD as the diameter is given by

(x – x_{1}) (x – x_{2}) + (y – y_{1}) (y – y_{2}) = 0

(x – 9) (x – 6) + (y – 3) (y – 6) = 0

Upon simplifying, we get

x^{2} – 9x – 6x + 54 + y^{2} – 3y – 6y + 18 = 0

x^{2} + y^{2} – 15x – 9y + 72 = 0

*∴The equation of the circle is x ^{2} + y^{2} – 15x – 9y + 72 = 0*

**4. Find the equation of the circle circumscribing the rectangle whose sides are x – 3y = 4, 3x + y = 22, x – 3y = 14 and 3x + y = 62.**

**Solution:**

Given:

The sides x – 3y = 4 …. (1)

3x + y = 22 … (2)

x – 3y = 14 …. (3)

3x + y = 62 … (4)

Let us assume A, B, C, D be the vertices of the square. On solving the lines, we get the coordinates as A = (7, 1)

B = (8, – 2)

C = (20, 2)

D = (19, 5)

We know that the equation of the circle with diagonal AC as the diameter is given by

(x – x_{1}) (x – x_{2}) + (y – y_{1}) (y – y_{2}) = 0

(x – 7) (x – 20) + (y – 1) (y – 2) = 0

Upon simplification, we get

x^{2} + y^{2} – 27x – 3y + 142 = 0

*∴The equation of the circle is x ^{2} + y^{2} – 27x – 3y + 142 = 0*

**5. Find the equation of the circle passing through the origin and the points where the line 3x + 4y = 12 meets the axes of coordinates.**

**Solution:**

Given:

The line 3x + 4y = 12

The value of x is 0 on meeting the y – axis. So,

3(0) + 4y = 12

4y = 12

y = 3

The point is A(0, 3)

The value of y is 0 on meeting the x-axis. So,

3x + 4(0) = 12

3x = 12

x = 4

The point is B(4, 0)

Since the circle passes through the origin and A and B,

So, AB is the diameter

We know that the equation of the circle with AB as the diameter is given by

(x – x_{1}) (x – x_{2}) + (y – y_{1}) (y – y_{2}) = 0

(x – 0) (x – 4) + (y – 3) (y – 0) = 0

x^{2} + y^{2} – 4x – 3y = 0

*∴The equation of the circle is x ^{2} + y^{2} – 4x – 3y = 0*

**6. Find the equation of the circle which passes through the origin and cuts off intercepts a and b, respectively, from the x and y-axes.**

**Solution:**

Since the circle has intercept ‘a’ from the x-axis, the circle must pass through (a, 0) and (-a, 0) as it already passes through the origin.

Since the circle has intercept ‘b’ from the x-axis, the circle must pass through (0, b) and (0, -b) as it already passes through the origin.

Let us assume the circle passes through points A(a,0) and B(0,b).

We know that the equation of the circle with AB as the diameter is given by

(x – x_{1}) (x – x_{2}) + (y – y_{1}) (y – y_{2}) = 0

(x – a) (x – 0) + (y – 0) (y – b) = 0

x^{2} + y^{2} + ax + by = 0 or x^{2} + y^{2} – ax – by = 0

*∴The equation of the circle is x ^{2} + y^{2} + ax + by = 0 or x^{2} + y^{2} – ax – by = 0*

### Also, access exercises of RD Sharma Solutions for Class 11 Maths Chapter 24 – The Circle

## Frequently Asked Questions on RD Sharma Solutions for Class 11 Maths Chapter 24

### What is the importance of using the RD Sharma Solutions for Class 11 Maths Chapter 24?

1. Subject experts have created the solutions carefully according to the latest CBSE guidelines.

2. The solutions PDF help students during their revisions.

3. Provide accurate answers to all the questions present in the textbook.

4. All the fundamentals and complex topics are explained in simple steps.

## Comments