# Class 11 RD Sharma Solutions – Chapter 24 The Circle – Exercise 24.2

• Last Updated : 08 May, 2021

### Question 1(i). Find the coordinates of the centre and radius of the circle x2 + y2 + 6x – 8y – 24 = 0.

Solution:

As we know that the general equation of circles is

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x2 + y2 + 2gx + 2fy + c = 0      …..(i)

Centre = (-g, -f)

Radius = Given equation of circle is x2 + y2 + 6x – 8y – 24 = 0

On comparing with eq(i), we get

Hence, g = 3, f = -4, c = -24

So,

Centre = (-3, 4)

Radius = ### Question 1(ii). Find the coordinates of the centre and radius of the circle 2x2 + 2y2 – 3x + 5y = 7.

Solution:

As we know that the general equation of circles is

x2 + y2 + 2gx + 2fy + c = 0      …..(i)

Centre = (-g, -f)

Radius = Given equation of circle is 2x2 + 2y2 – 3x + 5y – 7 = 0

x2 + y2 – 3/2x + 5/2y – 7/2 = 0

On comparing with eq(i), we get

Hence, g = -3/4, f = 5/4, c = -7/2

So, center = (3/4, -5/4)

Radius = ### Question 1(iii). Find the coordinates of the centre and radius of the circle 1/2(x2 + y2) + x cosθ + y sinθ – 4 = 0.

Solution:

As we know that the general equation of circles is

x2 + y2 + 2gx + 2fy + c = 0      …..(i)

Centre = (-g, -f)

Radius = Given equation of circle is 1/2(x2 + y2) + x cosθ + y sinθ – 4 = 0

⇒ x2 + y2 + 2x cosθ + 2sin θ – 8 = 0

On comparing with eq(i), we get

Hence, g = cos θ, f = sin θ, c = -8

So, centre = (-cos θ, -sinθ)

Radius = ### Question 1(iv). Find the coordinates of the centre and radius of the circle x2 + y2 – ax – by = 0.

Solution:

As we know that the general equation of circles is

x2 + y2 + 2gx + 2fy + c = 0      …..(i)

Centre = (-g, -f)

Radius = Given equation of circle is x2 + y2 – ax – by = 0

On comparing with eq(i), we get

Hence, g = -a/2, f = -b/2, c = 0

So, centre = (a/2, b/2)

Radius = Radius = ### Question 2(i). Find the equation of the circle passing through the points (5, 7), (8, 1), and (1, 3).

Solution:

Given that, the circle pass through points P(5, 7), Q(8, 1), and R(1, 3)

As we know that the general equation of circles is

x2 + y2 + 2gx + 2fy + c = 0      …..(i)

Since, P, Q, and R lies on eq(i)

So,

25 + 49 + 10g + 14f + c = 0 …….(ii)

64 + 1 + 16g + 2f + c = 0  …….(iii)

1 + 9 + 2g + 6f + c = 0  ……(iv)

Now on solving eq (ii), (iii), and (iv), we get,

g = -29/6, f = -19/6, c = 56/3

Now put all these values in eq(i), we get

The equation of circle is

x2 + y2 – 29/3x – 19/6y + 56/3 = 0

3(x2 + y2) – 29x – 19y + 56 = 0

### Question 2(ii). Find the equation of the circle passing through the points (1, 2), (3, -4), and (5, -6).

Solution:

Given that, the circle pass through points P(1, 2), Q(3, -4), and R(5, -6)

As we know that the general equation of circles is

x2 + y2 + 2gx + 2fy + c = 0      …..(i)

Since, P, Q, and R lies on eq(i)

So,

1 + 4 + 2g + 4f + c = 0          …..(ii)

9 + 16 + 6g – 8f + c = 0        …..(iii)

25 + 36 + 10g – 12f + c = 0      …..(iv)

Now on solving eq (ii), (iii), and (iv), we get,

g = -11, f = -2, and c = 25

Now put all these values in eq(i), we get

The equation of circle is

x2 + y2 – 22x – 4y + 25 = 0

### Question 2(iii). Find the equation of the circle passing through the points (5, -8), (-2, 9), and (2, 1).

Solution:

Given that, the circle pass through points P(5, -8), Q(-2, 9), and R(2, 1)

As we know that the general equation of circles is

x2 + y2 + 2gx + 2fy + c = 0      …..(i)

Since, P, Q, and R lies on eq(i)

So,

25 + 64 + 10g – 16f + c = 0     …..(ii)

4 + 81 – 4g + 18f + c = 0     …..(iii)

4 + 1 + 4g + 2f + c = 0      …..(iv)

Now on solving eq (ii), (iii), and (iv), we get,

g = 58, f = 24, and c = -285

Now put all these values in eq(i), we get

The equation of circle is

x2 + y2 + 116x + 48y – 285 = 0

### Question 2(iv). Find the equation of the circle passing through the points (0, 0), (-2, 1), and (-3, 2).

Given that, the circle pass through points P(0, 0), Q(-2, 1), and R(-3, 2)

As we know that the general equation of circles is

x2 + y2 + 2gx + 2fy + c = 0      …..(i)

Since, P, Q, and R lies on eq(i)

So,

0 + 00 + 0 + c = 0    …..(ii)

4 + 1 – 4x + 2y + c = 0    …..(iii)

9 + 4 – 6x + 4y + c = 0     …..(iv)

Now on solving eq (ii), (iii), and (iv), we get,

g = -3/2, f = -11/2, and c = 0

Now put all these values in eq(i), we get

The equation of circle is

x2 + y2 – 3x – 11y = 0

### Question 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:

It is given that, the circle passing through P(3, -2) and Q(-2, 0) and having its centre on 2x – y = 3.

As we know that the general equation of circles is

x2 + y2 + 2gx + 2fy + c = 0      …..(i)

Since the circle passes through (3, -2) and also (-2, 0)

So,

9 + 4 + 6g – 4f + c = 0    …..(ii)

4 + 0 – 4g + 0 + c = 0    …..(iii)

Also, the centre of the circle lies on 2x – y = 3

-2g + f = 3   …..(iv)

Now on solving eq (ii), (iii), and (iv), we get,

g = 3/2, f = 6, and c = 2

Now put all these values in eq(i), we get

The equation of circle is

x2 + y2 + 3x + 12y + 2 = 0

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

Solution:

It is given that, the circle passing through P(3, 7) and Q(5, 5) and having its centre on x – 4y = 1

As we know that the general equation of circles is

x2 + y2 + 2gx + 2fy + c = 0      …..(i)

Since the circle passes through P and Q

so,

9 + 49 + 7g + 14f + c = 0      …..(ii)

25 + 25 + 10g + 10f + c = 0     …..(iii)

Also, the centre of the circle lies on  x – 4y = 1

so, -g + 4f = 1     …..(iv)

Now on solving eq (ii), (iii), and (iv), we get,

g = 3, f = 1, and c = -90

Now put all these values in eq(i), we get

The equation of circle is

x2 + y2+ 6x + 2y – 90 = 0

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

Solution:

Given that P (3, -2), Q(1, 0), R(-1, -2), and S(1,-4)

As we know that the general equation of circles is

x2 + y2 + 2gx + 2fy + c = 0      …..(i)

Since the circle passes through P, Q and R

So,

9 + 4 + 6g – 4f + c = 0      …..(ii)

1 + 0 + 2g – 0 + c = 0      …..(iii)

1 + 4 – 2g – 4f + c = 0      …..(iv)

Now on solving eq (ii), (iii), and (iv), we get,

g = -1, f = 2 and c = 1

Now put all these values in eq(i), we get

The equation of circle is

x2 + y2 – 2x + 4y + 1 = 0      …..(v)

Here, we clearly see that point S(1,-4) satisfy eq(v)

Hence, points P, Q, R, and S are concyclic

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

Solution:

As we know that the general equation of circles is

x2 + y2 + 2gx + 2fy + c = 0      …..(i)

Centre = (-g, -f)

Radius = Therefore, P(5, 5), Q(6, 4), and R(-2, 4) lie on eq(i),

So,

25 + 25 + 10g + 10f + c = 0      …..(ii)

36 + 16 + 12g + 8f + c = 0      …..(iii)

4 + 16 + 4g + 8f + c = 0      …..(iv)

Now on solving eq (ii), (iii), and (iv), we get,

g = -2, f = -1, c = -20

Now put all these values in eq(i), we get

The equation of circle is

x2 + y2 – 4x – 2y – 20 = 0     …..(v)

Here, we clearly see that point S(7, 1)

Hence, points P, Q, R, and S are concyclic

Now, the centre = (-g, -f) = (2, 1)

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

Solution:

The equations of lines are

x + y = -3      …..(i)

x – y = -1      …..(ii)

x = 3      …..(iii)

Let us considered A, B, and C are the point of intersection of lines (i), (ii), and (iii)

So, Point A(-2,-1), B(3, 4), and C(3,-6)

As we know that the general equation of circles is

x2 + y2 + 2gx + 2fy + c = 0      …..(iv)

So, the circle circumscribing the ∆ABC

4 + 1 – 4g – 2f + c = 0      …..(v)

9 + 16 + 6g + Bf + c = 0      …..(vi)

9 + 36 + 6g – 12f + c = 0      …..(vii)

Now on solving eq (v), (vi), and (vii), we get,

g = -3, f = 1, c = -15

Now put all these values in eq(iv), we get

The equation of circle is

x2 + y2 – 6x + 2y – 15 = 0

### Question 7(ii). Find the equation of the circle which circumscribes the triangle formed by the lines 2x + y – 3 = 0, x + y – 1 = 0, and 3x + 2y – 5 = 0.

Solution:

The equations of lines are

2x + y = 3     …..(i)

x + y = 1     …..(ii)

3x + 3y = 5     …..(iii)

Let us considered A, B, and C are the point of intersection of lines (i), (ii), and (iii)

So, A = (2, -1)

B = (3, -2)

C = (1, 1)

As we know that the general equation of circles is

x2 + y2 + 2gx + 2fy + c = 0      …..(iv)

So, the circle circumscribing the ∆ABC

4 + 1 + 4g – 2f + c = 0     …..(v)

9 + 4 + 6g – 4f + c = 0     …..(vi)

1 + 1 + 2g + 2f + c = 0     …..(vii)

Now on solving eq (v), (vi), and (vii), we get,

g = -13/2, f = -5/2, and c = 16

Now put all these values in eq(iv), we get

The equation of circle is

x2 + y2 – 13x – 5y + 16 = 0

### Question 7(iii). Find the equation of the circle which circumscribes the triangle formed by the lines x + y = 2, 3x – 4y = 6, and x – y = 0.

Solution:

The equations of lines are

x + y = 2      …..(i)

3x – 4y = 6      …..(ii)

x – y = 0      …..(iii)

Let us considered A, B, and C are the point of intersection of lines (i), (ii), and (iii)

So, A = (2, 0)

B = (-6, -6)

C = (1, 1)

As we know that the general equation of circles is

x2 + y2 + 2gx + 2fy + c = 0      …..(iv)

So, the circle circumscribing the ∆ABC

4 + 4g + c = 0     …..(v)

36 + 36 – 12g – 12f + c = 0     …..(vi)

1 + 1 + 2g + 2f + c = 0     …..(vii)

Now on solving eq (v), (vi), and (vii), we get,

g = 2, f = 3, and c = -12

Now put all these values in eq(iv), we get

The equation of circle is

x2 + y2 + 4x + 6y – 12 = 0

### Question 7(iv). Find the equation of the circle which circumscribes the triangle formed by the lines y = x + 2, 3y = 4x, and 2y = 3x.

Solution:

The equations of lines are

y = x + 2      …..(i)

3y = 4x      …..(ii)

2y = 3x      …..(iii)

Let us considered A, B, and C are the point of intersection of lines (i), (ii), and (iii)

So, A = (6, 8)

B = (4, 6)

C = (0, 0)

As we know that the general equation of circles is

x2 + y2 + 2gx + 2fy + c = 0      …..(iv)

So, the circle circumscribing the ∆ABC

12g + 16f + c = -100      …..(v)

8g + 12f + c = -52      …..(vi)

c = 0      …..(vii)

Now on solving eq (v), (vi), and (vii), we get,

f = 11 and g = -23

Now put all these values in eq(iv), we get

The equation of circle is

x2 + y2 – 46x + 22y = 0.

### Question 8. Prove that the centre of the three circles x2 + y2 – 4x – 6y – 12 = 0, x2 + y2 + 2x + 4y – 10 = 0, and x2 + y2 -10x – 16y – 1 = 0 are collinear.

Solution:

The given equation of circles are,

x2 + y2 – 4x – 6y – 12 = 0  …….(i)

x2 + y2 + 2x + 4y – 10 = 0  …….(ii)

x2 + y2 – 10x – 16y – 1 = 0  …….(iii)

Let O1, O2 and O3 are the centres of (i), (ii), and (iii)

O1 = (-g, -f) = (2, 3)

O2 = (-g, -f) = (-1, -2)

O3 = (-g, -f) = (5, 8)

O1, O2 and O3 will be collinear if ar(∆ O1O2O3) = 0  R2 ⇒ R2 ⇒ -R1

R3 ⇒ R3 ⇒ -R1 O1, O2, and O3 are collinear

### Question 9. Prove that the radii of the circles x2 + y2 = 1, x2 + y2 – 2x – 6y – 6 = 0, and x2 + y2 – 4x – 12y – 9 = 0 are in A.P.

Solution:

The given equation of circles are,

x2 + y2 = 1 ———–(i)

x2 + y2 – 2x – 6y – 6 = 0 ———-(ii)

x2 + y2 – 4x – 12y – 9 = 0 ———-(iii)

Let us considered R1, R2, and R3 are the radii of (i), (ii), and (iii)

So, R1 = 1

R2 = R3 = As we know that if a, b, c are in AP, then b = a + b/2

so, a = 1, b = 4, c = 7, b = 1 + 7/2 = 4

Therefore 1, 4, 7 are in AP.

Hence, the radius of the three circles lie in AP.

### Question 10. Find the equation of the circle which passes through the origin and cuts off chord of length 4 and 6 on the positive side of the x-axis and y-axis respectively.

Solution:

It is given that a circle that passes through origin O(0, 0) and

cut off on intercepts of length 4 on x-axis and 6 on y-axis.

So, OA = 4

OB = 6

Let us assume C be the centre of the circle and

CM and CN are perpendicular line drawn on OA and OB

so, the coordinate of A = (4, 0) and B = (0, 6)

The coordinates of M = (2, 0) and N = (0, 3)

And the coordinates of C = (2, 3)

Now in ∆OCM,

Using Pythagoras theorem

OC2 = OM2 + CM2

= 22 + 32

= 4 + 9

OC = √13

Hence, the required circle is

(x – 2)2 + (y – 3)2 = 13

x2 + y2 – 4x – 6y = 0

### Question 11. Find the equation of the circle concentric with circles x2 + y2 – 6x + 12y + 15 = 0 and double of its area.

Solution:

The given equation of circle is

x2 + y2 – 6x + 12y + 15 = 0 …….(i)

so, centre = (-g, -f) = (3, 6)

radius  = Now, the required equation of circle in concentric with (i)

that means both have same centre (3,-6)

The area of required circle = 2 * Area of (i)

πR2 = 2 * π(√30)2

R2 = 60

R = 2√15

Hence, the required circle is

(x – 3)2 + (y + 6)2 = 60

x2 + y2 – 6x + 12y – 15 = 0

### Question 12. Find the equation of the circle which passes through the points (1, 1), (2, 2), and whose radius is 1. Show that there are two such circles.

Solution:

As we know that the general equation of circles is

x2 + y2 + 2gx + 2fy + c = 0      …..(i)

It is given that the points P(1, 1) and Q(2, 2) passes throught eq(1)

So,

1 + 1 + 2g + 2f + c = 0     …..(ii)

4 + 4 + 4g + 4f + c = 0     …..(iii)

Its is gievn that the radius = 1 ⇒ g2 + f2 – c = 1   ……(iv)

Now from eq(ii) and (iii), we get

g + f + c/2 = -1

g + f + c/4 = -2

Now on subtracting both the equations, we get

c = 4

g + f = -3  …….(v)

Now on solving eq(v) and (vi), we get

g = -1 or -2 and f = -2 or -1

Hence, the required circle is

x2 + y2 – 2x – 4y + 4 = 0

or

x2 + y2 – 4x – 2y + 4 = 0

### Question 13. Find the equation of the circle concentric with x2 + y2 – 4x – 6x – 6y – 3 = 0 and which touches the y-axis.

Solution:

The given equation of circle is

x2 + y2 – 4x – 6y – 3 = 0   …..(i)

so, centre = (-g, -f) = (2, 3)

The required circle is concentric with eq(i)

so, both have centre(2, 3)

Also, the required circle touches y-axis at A.

So, CA = radius = 2

Hence, the required circle is

(x – 2)2 + (y – 3)2 = 4

x2 + y2 – 4x – 6y + 9 = 0

### Question 14. If a circle passes through the point (0, 0), (a, 0), (0, b), then find the coordinates of its centre.

Solution: From the given figure, CA, CO, and AB  are the equal radii of the circle

So,

CA = CO = CB = r

Also, OCA is an isosceles triangle, and CM is the perpendicular bisector to the OA.

Hence  OM = a/2

Similarly, CN is the perpendicular bisector to the OB

So, ON = b/2

from the above figure, it is clear that

OM = x = a/2

ON = y = b/2

Hence the centre of the circle is c(a/2, b/2)

### Question 15. Find the equation of the circle which passes through the point (2, 3) and (4, 5), and the centre lies on the straight line y – 4x + 3 = 0.

Solution:

Its is given that a circle which passes through the point P(2, 3) and Q(4, 5) and

the centre lies on the straight line y – 4x + 3 = 0.

As we know that the general equation of circles is

x2 + y2 + 2gx + 2fy + c = 0      …..(i)

Since the circle passes through P, and Q

So,

13 + 14g + 6f + c = 0    …..(ii)

41 + 8g + 10f + c = 0    …..(iii)

Centre (-g, -f) lies on y – 4x + 3 = 0

So, -f + 4g = -3    …..(iv)

Now on subtracting eq(ii) from (iii), we get

28 + 4g + 4f = 0    …..(v)

On solving eq(iv) and (v) we get,

f = -5  and g = -2

Now put all these values in eq(i), we get

The equation of circle is

41 – 16 – 50 + c = 0

c = 25

Hence,  the required  equation of the circle is,

x2 + y2 – 4x – 10y + 25 = 0

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