Question

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

Answer 1

Answer: 816

Normal are
$y+2x=\sqrt{11}+7\sqrt{7}$
$2y+x=2\sqrt{11}+6\sqrt{7}$
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 $\sqrt{11}y-3x=\frac{5\sqrt{77}}{3}+11$
Radius will be $\perp$ distance of tangent from center
i.e. $4\sqrt{\frac{7}{5}}$
Now $(5h-8k{)}^2+5r^2=816$