Question

If $y=x$ be the tangent to the circle $x^2+y^2+2gx+2fy+c=0$ at ponit $P$ such that the distance of $P$ from origin is $4\sqrt{2},$ then the value of $c$ is

Answer 1

$x^2+y^2+2gx+2fy+c=0$
Let $r$ be the radius and $Q(-g,-f)$ be the centre of the circle.
$r=\sqrt{g^2+f^2-c}=PQ$


From figure,
$O{Q}^2=P{Q}^2+P{O}^2$
$\Rightarrow (\sqrt{g^2+f^2}{)}^2=g^2+f^2-c+(4\sqrt{2}{)}^2$
$\Rightarrow g^2+f^2=g^2+f^2-c+32$
$\Rightarrow c=32$