If the circle $x^2+y^2+2gx+2fy+c=0$ is touched by $y=x$ at P such that  $OP=6\sqrt{2},$ then the value of c is

The equation of the line y = x in distance form is $\frac{x}{\cos \theta }=\frac{y}{\sin \theta }=r,\text{ where }\theta =\frac{\pi }{4}$

For point $p$ we have $r=6\sqrt{2}.$

Therefore, coordinates of Pare given by 

$\frac{x}{\cos \frac{\pi }{4}}=-\frac{y}{\sin \frac{\pi }{4}}=6\sqrt{2}\Rightarrow x=6,y=6$

Since P(6, 6) lies on $x^2+y^2+2gx+2fy+c=0$

$\therefore 72+12(g+f)+c=0$

Since y = x touches the circle. Therefore, the equation

$2x^2+2x(g+f)+c=0$

$\Rightarrow 4(g+f{)}^2=8c\Rightarrow (g+f{)}^2=2c$

$\begin{array}{r}\left[12\right(g+f){]}^2=[-(c+72){]}^2\\ \Rightarrow 144(g+f{)}^2=(c+72{)}^2\\ \Rightarrow 144\left(2c\right)=(c+72{)}^2\\ \Rightarrow (c-72{)}^2=0\Rightarrow c=72\end{array}$