If the circles `x^2 + y^2 + 2gx + 2fy + c = 0` bisects `x^2+y^2+2g'x+2f'y + c' = 0` then the length of the common chord of these two circles is -

Options

A. `2sqrt(g^(2)+f^(2)-c)`
B. `2sqrt(g'^(2)+g'^(2)-c')`
C. `2sqrt(g^(2)+f^(2)+c)`
D. `2sqrt(g'^(2)+f'^(2)+c')`

Answer

B

Explanation

The length of the common chord of the two given circles is the length of the diameter of the circle
`x^(2)+y^(2)+2g'x+2f'y+c'=0`
Hence, the required length `=2sqrt(g'^(2)+f'^(2)-c')`