The line $x+y+2=0$ intersect the circle $x^2+y^2+4 x-$ $4 y-4=0$ in two points $A$ and $B$. Let $S \equiv…
The line $x+y+2=0$ intersect the circle $x^2+y^2+4 x-$ $4 y-4=0$ in two points $A$ and $B$. Let $S \equiv x^2+y^2+2 g x+$ $2 \mathrm{fy}+\mathrm{c}=0$ be a different circle passing through the points $A$ and $B$. If the distance of the centre of $S=0$ from $A B$ is $\sqrt{2}$, then $\mathrm{g}+\mathrm{f}+\mathrm{c}=$
$12$
$8$
$6$
$0$
Solution
Equation of circle(s) passing through intersection of line and circle is $x^2+y^2+4 x-4 y-4+\lambda(x+y+2)=0$ $x^2+y^2+(4+\lambda) x+(\lambda-4) y+(2 \lambda-4)=0$
But $x^2+y^2+2 g x+2 f y+c=0$ (Given)
$
\therefore g=\frac{4+\lambda}{2}, f=\frac{\lambda-4}{2}, c=2 \lambda-4
$
Given that distance of $(-g,-f)$ from $x+y+z=0$ is $\sqrt{2}$
$
\begin{aligned}
& \therefore\left|\frac{-g-f+2}{\sqrt{2}}\right|=\sqrt{2} \Rightarrow g+f=4 \\
& \frac{4+\lambda}{2}+\frac{\lambda-4}{2}=4 \Rightarrow \lambda=4 \Rightarrow g=4, f=0, c=4 \\
& \text { Now, } g+f+c=4+0+4=8
\end{aligned}
$