# Class 11 RD Sharma Solutions- Chapter 24 The Circle – Exercise 24.1 | Set 1

**Question 1. Find the equation of the circle with:**

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

**(ii) Centre (a, b) and radius √(a ^{2}+b^{2}).**

**(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.

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

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

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 √(a^{2}+b^{2}).

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 =√(a^{2}+b^{2})

Substituting the values in the above equation, we get

(x – p)^{2}+ (y – q)^{2}= r^{2}

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

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

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

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.

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

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

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.

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

Given, p = a, q = a, r = √2 a

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.

**Question 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

Using the standard equation formula,

(x – a)^{2}+ (y – b)^{2}= r^{2}… (1)

Let’s convert given circle’s equation into the standard form.

(x – 1)^{2}+ y^{2}= 4

(x – 1)^{2}+ (y – 0)^{2}= 22 ….. (2)

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

Using the standard equation formula,

(x – a)^{2}+ (y – b)^{2}= r^{2}…. (1)

Let’s convert given circle’s equation into the standard form.

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

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

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

Using the standard equation formula,

(x – a)^{2}+ (y – b)^{2}= r^{2}…. (1)

Let’s convert 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)

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

Using the standard equation formula,

(x – a)^{2}+ (y – b)^{2}= r^{2}…. (1)

Let’s convert 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)

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.

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

**Solution:**

Given, 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

r = 5

You 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}

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.

**Question 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’s find the points of intersection of the lines.

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

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

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

To find the radius of the circle.

We found, 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 = r^{2}

5 = r^{2}

r = √5

You 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}

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.

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

**Solution:**

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

A circle with centre (0, 6) and having radius 4.

You 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}

We found, p = 0, q = 6, r = 4

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.

**Question 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’s given that the circle has the radius 10 and has diameters 2x + y = 6 and 3x + 2y = 4.

The centre is the intersection point of the diameters.

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

A circle with centre (8, -10) and having radius 10.

You 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}

We found, p = 8, q = -10, r = 10

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.

**Question 7. Find the equation of the circle**

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

**(ii) Which touches 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)The circle touches the axes at the points (±6, 0) and (0, ±6), so, it has a centre (±6, ±6) and passes through the point (0, 6).

To find the radius of the circle.

We found, p = 6, q = 6

(x – 6)^{2}+ (y – 6)^{2}= r^{2}…. (1)

(0, 6) lies on Equation (1) so,

(0 – 6)^{2}+ (6 – 6)^{2}= r^{2}

36 + 0 = r^{2}

r = √36

r = 6

You 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}

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

(ii)The circle touches the x – axis at the points (±5, 0).

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

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

To find the radius of the circle.

We found, p = 5, q = a

(x – 5)^{2}+ (y – a)^{2}= r^{2}…. (1)

The above equation passes through (5, 0) so,

(5 – 5)^{2}+ (0 – 6)^{2}= r^{2}

0 + 36 = r^{2}

r = √36

r = 6

The circle has centre at (±5, ±6) and having radius 6 units.

Substituting the values in the standard 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.

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

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

Substituting the values in standard form we get,

(x – a)^{2}+ (y – a)^{2}= a^{2}… (1)

Equation (1) passes through P (2, 1)

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 1Case (i)

The centre at (5, 5) and having radius 5 units.

Substituting the values in the standard 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.Case (ii)

The centre at (1, 1) and having a radius 1 unit.

Substituting the values in the standard 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

(iv)Let’s assume the abscissa as ‘a’

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

To find the radius of the circle.

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.

Substituting the values in the standard 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.

**Question 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:**

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

Usining the perpendicular distance between a point and a straight line formula we get, 62/13.

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

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 = r2

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

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

**Solution:**

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

The centre of the circle is (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 circle with centre (-3, -3) and having radius 3.

Substituting the values in the standard 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

**Question 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’s 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.

Solving 2x – 3y + 4 = 0 and 3x + 4y – 5 = 0 we get the intersection point as ((-1/17),(22/17)).

To find radius, we use distance formula between the points ((-1/17),(22/17)) and (0, 0)=√485/17

You 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}

Substituting the values in the equation, we get

(x-(-1/17)^{2}+(y-(22/17))^{2}=(√485/17)^{2}