# Conic Sections – Exercise 11.1 – Class XI

1. Find the equation of the circle with centre (0, 2) and radius 2

Solution:

The equation of a circle with centre (h, k) and radius r is given as (x – h)2 + (y – k)2 = r2

It is given that centre (h, k) = (0, 2) and radius (r) = 2.

Therefore, the equation of the circle is

(x – 0)2 + (y – 2)2 = 22

⟹ x2 + y2 + 4 – 4 y = 4

⟹ x2 + y2 – 4y = 0

1. Find the equation of the circle with centre (–2, 3) and radius 4

Solution:

The equation of a circle with centre (h, k) and radius r is given as

(x – h)2 + (y – k)2 = r2

It is given that centre (h, k) = (–2, 3) and radius (r) = 4.

Therefore, the equation of the circle is

(x + 2)2 + (y – 3)2 = (4)2

⟹ x2 + 4x + 4 + y2 – 6y + 9 = 16

⟹ x2 + y2 + 4x – 6y – 3 = 0

1. Find the equation of the circle with centre(1/2, 1/4) and  radius(1/12)

Solution:

The equation of a circle with centre (h, k) and radius r is given as

(x – h)2 + (y – k)2 = r2

It is given that centre (h, k) =(1/2, 1/4) and radius (r) = (1/12)

Therefore, the equation of the circle is

(x – 1/2)2 + (y – 1/4)2 = (1/12)2

x2 – x + 1/4 + y2y/2 + 1/16 = 1/144

x2 – x + 1/4 + y2y/2 + 1/161/144 = 0

144x2 – 144x + 36 + 144y2 – 72y + 44 = 0

36x2 – 36x + 36y2 – 18y + 11 = 0

36x2 + 36y2 – 36x – 18y + 11=0

1. Find the equation of the circle with centre (1, 1) and radius √2.

Solution:

The equation of a circle with centre (h, k) and radius r is given as

(x – h)2 + (y – k)2 = r2

It is given that centre (h, k) = (1, 1) and radius (r) = √2.

Therefore, the equation of the circle is

(x – 1)2 + (y – 1)2 = (√2)2

x2 – 2x + 1+y2 – 2y + 1 = 2

x2 + y2 -2x – 2y =0

1. Find the equation of the circle with centre (-a, -b) and radius √(a2 – b2).

Solution:

The equation of a circle with centre (h, k) and radius r is given as

(x – h)2 + (y – k)2 = r2

It is given that centre (h, k) = (-a, -b) and radius (r) = √(a2 – b2).

Therefore, the equation of the circle is

(x + a)2 + (y + b)2 = (√(a2 – b2))2

x2 + 2ax + a2 +y2 + 2by + b2 = a2 – b2

x2 + y2 + 2ax + 2by + 2b2 = 0

1. Find the centre and radius of the circle (x + 5)2 + (y – 3)2 = 36

Solution:

The equation of the given circle is (x + 5)2 + (y – 3)2 = 36.

(x + 5)2 + (y – 3)2 = 36

⇒ {x – (–5)}2 + (y – 3)2 = 62, which is of the form (x – h)2 + (y – k)2 = r2, where h = – 5, k = 3, and r = 6.

Thus, the centre of the given circle is (–5, 3), while its radius is 6.

1. Find the centre and radius of the circle x2 + y2 – 4x – 8y – 45 = 0

Solution:

The equation of the given circle is x2 + y2 – 4x – 8y – 45 = 0.

x2 + y2 – 4x – 8y – 45 = 0

⇒ (x2 – 4x) + (y2 – 8y) = 45

⇒ {x2 – 2(x)(2) + 22} + {y2 – 2(y)(4)+ 42} – 4 –16 = 45

⇒ (x – 2)2 + (y –4)2 = 65

⇒ (x – 2)2 + (y – 4)2= 65 ,

Which is of the form (x – h)2 + (y – k)2 = r2, where h = 2, k = 4 and since r2 = 65, we have r = √65 ,

Thus, the centre of the given circle is (2, 4), while its radius is √65 .

1. Find the centre and radius of the circle x2 + y2 – 8x + 10y – 12 = 0

Solution:

The equation of the given circle is x2 + y2 – 8x + 10y – 12 = 0.

x2 + y2 – 8x + 10y – 12 = 0

⇒ (x2 – 8x) + (y2 + 10y) = 12

⇒ {x2 – 2(x)(4) + 42} + {y2 + 2(y)(5) + 52}– 16 – 25 = 12

⇒ (x – 4)2 + (y + 5)2 = 53

⟹ (x − 4)2 +{y − (−5)}2 = (√53)2 which is of the form (x – h)2 + (y – k)2 = r2, where h = 4, k = – 5 and r = √53

Thus, the centre of the given circle is (4, –5), while its radius is √53.

1. Find the centre and radius of the circle 2x2 + 2y2 – x = 0

Solution:

The equation of the given circle is 2x2 + 2y2 – x = 0.

2x2 + 2y – x = 0

(2x2 – x)+2y2 = 0

2[(x2x/2)+y2] = 0

{x2 – 2x(1/4)+(1/4)2 } + y2 – (1/4)2 = 0

(x – 1/4)2 + (y – 0)2 = (1/4)2

which is of the form (x – h)2 + (y – k)2 = r2, where h = 1/4, k = 0 and r = 1/4.

Thus, the centre of the given circle is (1/4, 0), while its radius is 1/4.

1. Find the equation of the circle passing through the points (4, 1) and (6, 5) and whose centre is on the line 4x + y = 16.

Solution:

Let the equation of the required circle be (x – h)2 + (y – k)2 = r2.

Since the circle passes through points (4, 1) and (6, 5),

(4 – h)2 + (1 – k)2 = r2 …………………. (1)

(6 – h)2 + (5 – k)2 = r2 …………………. (2)

Since the centre (h, k) of the circle lies on line 4x + y = 16,

4h + k = 16 ———————(3)

From equations (1) and (2), we obtain

(4 – h)2 + (1 – k)2 = (6 – h)2 + (5 – k)2

⇒ 16 – 8h + h2 + 1 – 2k + k2 = 36 – 12h + h2 + 25 – 10k + k2

⇒ 16 – 8h + 1 – 2k = 36 – 12h + 25 – 10k

⇒ 4h + 8k = 44

⇒ h + 2k = 11 ———————–(4)

On solving equations (3) and (4), we obtain h = 3 and k = 4.

On substituting the values of h and k in equation (1), we obtain (4 – 3)2 + (1 – 4)2 = r2

⇒ (1)2 + (– 3)2 = r2

⇒ 1 + 9 = r2

⇒ r2 = 10

⇒ r = √10

Thus, the equation of the required circle is

(x – 3)2 + ( y – 4)2 = (√10)2

x2 – 6x + 9 + y2 – 8y + 16 = 10

x2 – 6x + y2 – 8y + 15 = 0

1. Find the equation of the circle passing through the points (2, 3) and (–1, 1) and whose centre is on the line x – 3y – 11 = 0.

Solution:

Let the equation of the required circle be (x – h)2 + (y – k)2 = r2.

Since the circle passes through points (2, 3) and (–1, 1),

(2 – h)2 + (3 – k)2 = r2——————- (1)

(–1 – h)2 + (1 – k)2 = r2 ————–(2)

Since the centre (h, k) of the circle lies on line x – 3y – 11 = 0,

h – 3k = 11 ———–(3)

From equations (1) and (2), we obtain

(2 – h)2 + (3 – k)2 = (–1 – h)2 + (1 – k)2

⇒ 4 – 4h + h2 + 9 – 6k + k2 = 1 + 2h + h2 + 1 – 2k + k2

⇒ 4 – 4h + 9 – 6k = 1 + 2h + 1 – 2k

⇒ 6h + 4k = 11 ————(4)

On solving equations (3) and (4), we obtain h =7/2 and k =−5/2

On substituting the values of h and k in equation (1), we get

(2 – 7/2)2 + (3 + 5/2)2 = r2

(4-7/2)2 + (6+5/2)2 = r2

(-3/2)2 + (11/2)2 = r2

9/4+ 121/4 = r2

130/4 = r2

Thus the equation of the required circle is

(x – 7/2)2 + (y + 5/2)2 = 130/4

(2x-7/2)2 + (2y+5/2)2 = 130/4

4x2 – 28x + 49 + 4y2 + 20y + 25 = 130

4x2 + 4y2 – 28x + 20y – 56 = 0

4(x2 + y2 – 7x + 5y – 14) = 0

x2 + y2 – 7x + 5y – 14 = 0

1. Find the equation of the circle with radius 5 whose centre lies on x-axis and passes through the point (2, 3).

Solution:

Let the equation of the required circle be (x – h)2 + (y – k)2 = r2.

Since the radius of the circle is 5 and its centre lies on the x-axis, k = 0 and r = 5.

Now, the equation of the circle becomes (x – h)2 + y2 = 25.

It is given that the circle passes through point (2, 3).

( 2  – h)2 + 32 = 25

( 2  – h)2 = 25 – 9 = 16

2 – h = ±√16 = ±4

If 2 – h = 4 then h = -2

If 2 – h = -4 then h = 6

when h = -2 the equation of the circle becomes

(x + 2)2 + y2 = 25

x2 + 4x + 4 + y2 – 25 = 0

x2 + 4x + y2 – 21 = 0

When h = 6 the equation of the circle becomes

(x – 6)2 + y2 = 25

x2 – 12x + 36 + y2 – 25 = 0

x2 – 12x + y2 + 11 = 0

1. Find the equation of the circle passing through (0, 0) and making intercepts a and b on the coordinate axes.

solution:

Let the equation of the required circle be (x – h)2 + (y – k)2 = r2.

Since the centre of the circle passes through (0, 0),

(0 – h)2 + (0 – k)2 = r2

⇒ h2 + k2 = r2

The equation of the circle now becomes (x – h)2 + ( y- k)2 =h2 + k2

It is given that the circle makes intercepts a and b on the coordinate axes. This means that the circle passes through points (a, 0) and (0, b). Therefore,

(a – h)2 + (0 – k)2 = h2 + k2————- (1)

(0 – h)2 + (b – k)2 = h2 + k2————— (2)

From equation (1), we obtain a2 – 2ah + h2 + k2 = h2 + k2

⇒ a2 – 2ah = 0

⇒ a(a – 2h) = 0

⇒ a = 0 or (a – 2h) = 0

However, a ≠ 0; hence, (a – 2h) = 0 ⇒ h =a/2.

From equation (2), we obtain h2 + b2 – 2bk + k2 = h2 + k2

⇒ b2 – 2bk = 0

⇒ b(b – 2k) = 0

⇒ b = 0 or(b – 2k) = 0

However, b ≠ 0; hence, (b – 2k) = 0 ⇒ k =b/2.

Thus, the equation of the required circle is

(x – a/2)2 + (y – b/2 )2 = ( a/2)2 +( b/2)2

(2x-a/2)2 + (2y-b/2 )2 = [a2 +b2]/2

4x2 – 4ax +a2 + 4y2 – 4by + b2 = a2 + b2

4x2+ 4y2– 4ax– 4by = 0

x2+ y2– ax – by = 0

1. Find the equation of a circle with centre (2, 2) and passes through the point (4, 5).

Solution:

The centre of the circle is given as (h, k) = (2, 2).

Since the circle passes through point (4, 5), the radius (r) of the circle is the distance between the points (2, 2) and (4, 5).

r = √[(2-4)2 + (2-5)2] = √[(2)2 + (3)2] = √(4+9) = √13

Thhus the equation of the circle is

( x- h)2 + ( y-k)2 = r2

( x- 2)2 + ( y-2)2 = (√13)2

x2 – 4x + 4 +y2 -4y + 4 = 13

x2 + y2– 4x-4y-5 = 0

1. Does the point (–2.5, 3.5) lie inside, outside or on the circle x2 + y2 = 25?

Solution:

The equation of the given circle is x2 + y2 = 25.

x2 + y2 = 25

⇒ (x – 0)2 + (y – 0)2 = 52,which is of the form (x – h)2 + (y – k)2 = r2, where h = 0, k = 0, and r = 5.

∴ Centre = (0, 0) and radius = 5

Distance between point (–2.5, 3.5) and centre (0, 0)

=√[(-2.5-0)2+(3.5-0)2]

=√(6.25+12.25)

=√18.5

≈4.3 <5

Since the distance between point (–2.5, 3.5) and centre (0, 0) of the circle is less than the radius of the circle, point (–2.5, 3.5) lies inside the circle.