Two equal circles with their centres as x and y axis will possess the radical axis in the following form

Options

A. `ax-by-(a^(2)+b^(2))/(4)=0`
B. `2gx-2fy+g^(2)-f^(2)=0`
C. `g^(2)x+f^(2)y-g^(4)-f^(4)=0`
D. `2g^(2)x+2f^(2)y-g^(4)-f^(4)=0`

Answer

B

Explanation

Let `(x-a)^(2)+(y-0)^(2)=r^(2) and x^(2)+(y-b)^(2)=r^(2)`
be two equal circles with their centres on x and y axes respectively. Then, their radical axis is given by
`-2ax+a^(2)+2by-b^(2)=0`
`rArr 2ax-2by+b^(2)-a^(2)=0`
Clearly, is of the form `2gx-2fy+g^(2)-f^(2)=0`.