If $b > a$ then prove that the length of the tangent drawn from any point on the circle $x^{2} + y^{2} + 2 g x + 2 f y + a = 0$ to the circle $x^{2} + y^{2} + 2 g x + 2 f y + b = 0$ is $b - a$ .

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Solution

If $P$ be $(h, k)$ on $1 s t$ , then
$h^{2} + k^{2} + 2 g h + 2 f y + a = 0 . . . . (1)$
$∴ t^{2} = S^{'} = h^{2} + k^{2} + 2 g h + 2 f y + b = b - a$ ...... by $(1)$ .

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