Question

Let the lines $y+2 x=\sqrt{11}+7 \sqrt{7}$ and $2 y + x =2 \sqrt{11}+6 \sqrt{7}$ be normal to a circle $C:(x-h)^{2}+(y-k)^{2}=r^{2}$. If the line $\sqrt{11} y-3 x=\frac{5 \sqrt{77}}{3}+11$ is tangent to the circle $C$, then the value of $(5 h-8 k)^{2}+5 r^{2}$ is equal to ______

Answer 1

Normal are
$y+2 x=\sqrt{11}+7 \sqrt{7} $
$2 y+x=2 \sqrt{11}+6 \sqrt{7}$
Center of the circle is point of intersection of normals i.e.
$\left(\frac{8 \sqrt{7}}{3}, \sqrt{11}+\frac{5 \sqrt{7}}{3}\right)$
Tangent is $\sqrt{11} y -3 x =\frac{5 \sqrt{77}}{3}+11$
Radius will be $\perp$ distance of tangent from center
i.e. $4 \sqrt{\frac{7}{5}}$
Now $(5 h -8 k )^{2}+5 r ^{2}=816$