Question

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 prove that $c=72$.

Answer 1

The point $P$ lies on $x-y=0$
or $\frac{x-0}{\cos {45}^{\circ }}=\frac{y-0}{\sin {45}^{\circ }}=r=OP=6\sqrt{2}$
$P$ is $(6,6)$. It also lies on circle
$72+12(g+f)+c=0....\left(1\right)$
Since $y=x$ touches the given circle, its intersection with given circle will have equal roots.
$\therefore 2x^2+2x(g+f)+c=0$ has equal roots.
$\therefore (g+f{)}^2=2c.....(2)$
Note : You will get the same result if you apply $p=r$.
Eliminating $(g+f)$ between $\left(1\right)$ and $\left(2\right)$, we get
$\frac{[-(c+72){]}^2}{144}=2c$
or $(c+72{)}^2-4\times 72\times c=0$ or $(c-72{)}^2=0$
$\therefore c=72[\because (x+y{)}^2-4xy=(x-y{)}^2]$.