Chapter 11: Conic Sections

The circle touches both the x and y axes in the first quadrant and the radius is a.

For a circle of radius a, the centre is (a, a).

The equation of a circle having centre (h, k), having radius as r units, is

(x – h)² + (y – k)² = r²

Therefore, the equation of the circle becomes (x – a)² + (y – a)² = a²

x² – 2ax + a² + y² – 2ay + a² – a² = 0

x² – 2ax + y² – 2ay + a² = 0

The equation of a circle having centre (h, k), having radius as r units, is

(x – h)² + (y – k)² = r²

Putting the values of given co-ordinates in the above expression,

(0, 0)

(0 – h)² + (0 – k)² = r²

h² + k² = r²

(a, 0)

(a – h)² + (0 – k)² = r²

a² + h² – 2ah + k² = r² —— (1)

(0, b)

(0 – h)² + (b – k)² = r²

h² + b² + k² – 2bk = r² ——- (2)

On solving equations (1) & (2), respectively,

a (a – 2h) = 0

b (b – 2k) = 0

So, a = 0 or 2h

b = 0 or 2k respectively.

Since the circle passes through the centre (0, 0), so the co-ordinates are

a = 2h, h = a/2

b = 2k, k = b/2

The co-ordinates of the centre are (a/2, b/2)

Since the circle has a centre (1, 2) and also touches x-axis.

Radius of the circle is, r = 2

The equation of a circle having centre (h, k), having radius as r units, is

(x – h)² + (y – k)² = r²

So, the equation of the required circle is:

(x – 1)² + (y – 2)² = 2²

⇒x² – 2x + 1 + y² – 4y + 4 = 4

⇒ x² + y² – 2x – 4y + 1 = 0

The equation of the circle is x² + y² – 2x – 4y + 1 = 0.

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

Given equation of the circle,

x² – 4x + y² – 6y + 11 = 0

x² – 4x + 4 + y² – 6y + 9 +11 – 13 = 0

the above equation can be written as

x² – 2 (2) x + 2² + y² – 2 (3) y + 3² +11 – 13 = 0

on simplifying we get

(x – 2)² + (y – 3)² = 2

(x – 2)² + (y – 3)² = (√2)²

Since, the equation of a circle having centre (h, k), having radius as r units, is

(x – h)² + (y – k)² = r²

We have centre = (2, 3)

The centre point is the mid-point of the two ends of the diameter of a circle.

Let the points be (p, q)

(p + 3)/2 = 2 and (q + 4)/2 = 3

p + 3 = 4 & q + 4 = 6

p = 1 & q = 2

Hence, the other ends of the diameter are (1, 2).

Solving the given equations,

3x + y = 14 ……….1

2x + 5y = 18 …………..2

Multiplying the first equation by 5, we get

15x + 5y = 70…….3

2x + 5y = 18……..4

Subtract equation 4 from 3 we get

13 x = 52,

Therefore x = 4

Substituting x = 4, in equation 1, we get

3 (4) + y = 14

y = 14 – 12 = 2

So, the point of intersection is (4, 2)

Since, the equation of a circle having centre (h, k), having radius as r units, is

(x – h)² + (y – k)² = r²

Putting the values of (4, 2) and centre co-ordinates (1,-2) in the above expression, we get

(4 – 1)² + (2 – (-2))² = r²

3² + 4² = r²

r² = 9 + 16 = 25

r = 5 units

So, the expression is

(x – 1)² + (y – (-2))² = 5²

Expanding the above equation we get

x² – 2x + 1 + (y + 2)² = 25

x² – 2x + 1 + y² + 4y + 4 = 25

x² – 2x + y² + 4y – 20 = 0

Hence the required expression is x² – 2x + y² + 4y – 20 = 0.

Hence, the required value of k is 8.

Given equation of the circle is

x² – 6x + y² + 12y + 15 = 0

The above equation can be written as

x² – 2 (3) x + 3² + y² + 2 (6) y + 6² + 15 – 9 + 36 = 0

(x – 3)² + (y-(-6))² – 30 = 0

(x – 3)² + (y-(-6))² = (√30)²

Since, the equation of a circle having centre (h, k), having radius as r units, is

(x – h)² + (y – k)² = r²

Centre = (3,-6)

Area of inner circle = πr² = 22/7 × 30 = 30 π units square

Area of outer circle = 2 × 30 π = 60 π units square

So, πr² = 60 π

r² = 60

Equation of outer circle is,

(x – 3)² + (y – (-6))² = (√60)²

x² – 6x + 9 + y² – 12 y + 36 = 60

x² – 6x + y² +12y +45 – 60 = 0

x² – 6x + y² + 12y – 15 = 0

Hence, the required equation of the circle is x² – 6x + y² + 12y – 15 = 0.