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

• Last Updated : 13 Jan, 2021

Question 1. Find the equation of the circle with:

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

(ii) Centre (a, b) and radius √(a2+b2).

(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 = r2
Given, p = -2, q = 3, r = 4
Substituting the values in the above equation, we get
(x – p)2 + (y – q)2 = r2
(x – (-2))2 + (y – 3)2 = 42
(x + 2)2 + (y – 3)2 = 16
x2 + 4x + 4 + y2 – 6y + 9 = 16
x2 + y2 + 4x – 6y – 3 = 0
∴ The equation of the circle is x2 + y2 + 4x – 6y – 3 = 0

(ii) Centre (a, b) and radius √(a2+b2).
The equation of the circle with centre (p, q) and radius ‘r’ is (x – p)2 + (y – q)2 = r2
Where, p = a, q = b, r =√(a2+b2)
Substituting the values in the above equation, we get
(x – p)2 + (y – q)2 = r2
(x – a)2 + (y – b)2 = (√(a2+b2))2
x2 – 2ax + a2 + y2 – 2by + b2 = a2 + b2
x2 + y2 – 2ax – 2by = 0
∴ The equation of the circle is x2 + y2
– 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 = r2
Given, p = 0, q = -1, r = 1
Substituting the values in the above equation, we get
(x – p)2 + (y – q)2 = r2
(x – 0)2 + (y – (-1))2 = 12
(x – 0)2 + (y + 1)2 = 1
x2 + y2 + 2y + 1 = 1
x2 + y2 + 2y = 0
∴ The equation of the circle is x2 + y2 + 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 = r2
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 = r2
(x – a cosα)2 + (y – a sinα)2 = a2
x2– (2acosα)x + a2cos2α + y2– (2asinα)y + a2sin2α = a2
We know that sin2θ + cos2θ = 1
So,
x2– (2acosα)x + y2– 2asinαy + a2 = a2
x2 + y2 – (2acosα)x – (2asinα)y = 0
∴ The equation of the circle is x2 + y2 – (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 = r2
Given, p = a, q = a, r = √2 a
Substituting the values in the above equation, we get
(x – p)2 + (y – q)2 = r2
(x – a)2 + (y – a)2 = (√2 a)2
x2 – 2ax + a2 + y2 – 2ay + a2 = 2a2
x2 + y2 – 2ax – 2ay = 0
∴ The equation of the circle is x2 + y2 – 2ax – 2ay = 0.

Question 2. Find the centre and radius of each of the following circles:

(i) (x – 1)2 + y2 = 4

(ii) (x + 5)2 + (y + 1)2 = 9

(iii) x2 + y2 – 4x + 6y = 5

(iv) x2 + y2 – x + 2y – 3 = 0

Solution:

(i) (x – 1)2 + y2 = 4
Using the standard equation formula,
(x – a)2 + (y – b)2 = r2 … (1)
Let’s convert given circle’s equation into the standard form.
(x – 1)2 + y2 = 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 = r2 …. (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 = 32 …. (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) x2 + y2 – 4x + 6y = 5
Using the standard equation formula,
(x – a)2 + (y – b)2 = r2 …. (1)
Let’s convert given circle’s equation into the standard form.
x2 + y2 – 4x + 6y = 5
(x2 – 4x + 4) + (y2 + 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) x2 + y2 – x + 2y – 3 = 0
Using the standard equation formula,
(x – a)2 + (y – b)2 = r2 …. (1)
Let’s convert given circle’s equation into the standard form.
x2 + y2 – x + 2y – 3 = 0
(x2 – x + ¼) + (y2 + 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 = r2
(x – 1)2 + (y – 2)2 = r2
It passes through the point (4, 6)
(4 – 1)2 + (6 – 2)2 = r2
32 + 42 = r2
9 + 16 = r2
25 = r2
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 = r2
Substituting the values in the above equation, we get
(x – 1)2 + (y – 2)2 = 52
x2 – 2x + 1 + y2 – 4y + 4 = 25
x2 + y2 – 2x – 4y – 20 = 0.
∴ The equation of the circle is x2 + y2 – 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 = r2 …. (1)
Equation (1) passes through (0, 0)
So, (0 + 2)2 + (0 – 1)2 = r2
4 + 1 = r2
5 = r2
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 = r2
Substituting the values in the above equation, we get
(x – (-2))2 + (y – 1)2 = (√5)2
(x + 2)2 + (y – 1)2 = 5
x2 + 4x + 4 + y2 – 2y + 1 = 5
x2 + y2 + 4x – 2y = 0
∴ The equation of the circle is x2 + y2 + 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 = r2
We found, p = 0, q = 6, r = 4
Substituting the values in the equation, we get
(x – 0)2 + (y – 6)2 = 42
x2 + y2 – 12y + 36 = 16
x2 + y2 – 12y + 20 = 0.
∴ The equation of the circle is x2 + y2 – 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 = r2
We found, p = 8, q = -10, r = 10
Substituting the values in the equation, we get
(x – 8)2 + (y – (-10))2 = 102
(x – 8)2 + (y + 10)2 = 100
x2 – 16x + 64 + y2 + 20y + 100 = 100
x2 + y2 – 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 = r2 …. (1)
(0, 6) lies on Equation (1) so,
(0 – 6)2 + (6 – 6)2 = r2
36 + 0 = r2
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 = r2
Substituting the values in the equation, we get
(x ± 6)2 + (y ± 6)2 = (6)2
x2 ± 12x + 36 + y2 ± 12y + 36 = 36
x2 + y2 ± 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 = r2 …. (1)
The above equation passes through (5, 0) so,
(5 – 5)2 + (0 – 6)2 = r2
0 + 36 = r2
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
x2 ± 10x + 25 + y2 ± 12y + 36 = 36
x2 + y2 ± 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 = a2 … (1)
Equation (1) passes through P (2, 1)
Substituting the values we get,
(2 – a)2 + (1 – a)2 = a2
4 – 4a + a2 + 1 – 2a + a2 = a2
5 – 6a + a2 = 0
(a – 5) (a – 1) = 0
So, a = 5 or 1
Case (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 = 52
x2 – 10x + 25 + y2 – 10y + 25 = 25
x2 + y2 – 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 = 12
x2 – 2x + 1 + y2 – 2y + 1 = 1
x2 + y2 – 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,
172 = a2 + (-15)2
289 = a2 + 225
a2 = 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 = 172
x2 ± 16x + 64 + y2 – 30y + 225 = 289
x2 + y2 ± 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
x2 – 6x + 9 + y2 – 8y + 16 = 3844/169
169x2 + 169y2 – 1014x – 1352y + 4225 = 3844
169x2 + 169y2 – 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 = 32
(x + 3)2 + (y + 3)2 = 9
x2 + 6x + 9 + y2 + 6y + 9 = 9
x2 + y2 + 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 = r2
Substituting the values in the equation, we get
(x-(-1/17)2+(y-(22/17))2=(√485/17)2

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