The equation of the circle on the common chord of the circles `(x-a)^(2)+y^(2)=a^(2)` and `x^(2)+(y-b)^(2)=b^(2)` as diameter, is

Options

A. `x^(2)+y^(2)=2ab(bx+ay)`
B. `x^(2)+y^(2)=bx+ay)`
C. `(a^(2)+b^(2))(x^(2)+y^(2))=2ab(bx-ay)`
D. `(a^(2)+b^(2))(x^(2)+y^(2))=2(bx+ay)`

Answer

C

Explanation

The equation of the common chord of the circles
`(x-a)^(2)+y^(2)=a^(2) and x^(2)+(y+b)^(2)=b^(2)` is
`ax+by=0 ` ...(i)
The equation of the required circle is
`{(x-a)^(2)+y^(2)-a^(2)} + lambda(ax+by)=0`
Since (i) is a diameter of circle (ii).
`:. a{-((a lambda-2a))/(2)}+b (-(lambda b)/(2))=0 rArr lambda = (2^(2))/(a^(2)+b^(2))`
Putting the value of `lambda` in (ii), the equation of the required circle is
`(a^(2)+b^(2))(x^(2)+y^(2))=2ab(bx-ay)`