Question

The equation of a circle which is coaxial with the circles

x² + y² + 4x + 2y + 1 = 0 and x² + y² – x + 3y – $\frac { 3 } { 2 }$ = 0 and having its centre on the radical axis of these circles, is –

• A. 4x² + 4y² + 6x + 10y – 1= 0
• B. 4x² + y² + 6x + 10y – 1 = 0
• C. x² + y² + 6x + 10y – 1 = 0
• D. 4x² + 4y² – 6x – 10y – 1 = 0

Answer 1

Answer: (A)

4x² + 4y² + 6x + 10y – 1= 0

Answer 2

The given circles are

S₁ ŗ x² + y² + 4x + 2y + 1 = 0 … (1)

and S₂ ŗ x² + y² – x + 3y – 3/2 = 0 … (2)

The radical axis of these two circles is

S₁ – S₂ = 0

Ž 5x – y + $\frac { 5 } { 2 }$ = 0

Ž 10x – 2y + 5 = 0 … (3)

The equation of the family of circles coaxial with the circles (1) and (2) isS₁ + l (S₁ – S₂) = 0

Ž (x² + y² + 4x + 2y + 1) + l (10x – 2y + 5) = 0 … (4)

Ž x² + y² + 2x (2x + 5l) + 2(1 – l)y + 1 + 5l = 0

The coordinates of the centre are (–(2 + 5l), (1 – l))

This circle will have its centre on the radical axis of the given circles i.e., on equation (3), if

–10(2 + 5l) + 2(1 – l) + 5 = 0

Ž l = – $\frac { 1 } { 4 }$

On putting l = – $\frac { 1 } { 4 }$ in equation (4), we get

4x² + 4y² + 6x + 10y – 1 = 0

as the equation of the required circle.