The locus of the centre of the circle which bisects the circumferences of the circles `x^2 + y^2 = 4 & x^2 + y^2-2x + 6y + 1 =0` is :

Options

A. `2x - 6y - 15 = 0`
B. `2x +6y +15 = 0`
C. `2x - 6y +15 = 0`
D. `2x +6y - 15 = 0`

Answer

A

Explanation

Let required circle be `x^(2) + y^(2) + 2gx +2fy +c =0`.
Hence, common chord with `x^(2) + y^(2) -4 =0` is
`2gx +2fy +c +4 =0`
This is diameter of circle `x^(2) + y^(2) =4` hence `c =-1.` Now, again common chord with other circle is:
`2x(g+1) + 2y (f-3) +(c-1) =0`
This is diameter of `x^(2) + y^(2) -2x +6y +1 =0`.
`:. 2(g+1) -6 (f-3) -5 =0`
or `2g - 6f +15 =0`
Locus is `2x -6y -15 =0`, which is straight line.