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

Normals are$\begin{array}{r}y+2x=\sqrt{11}+7\sqrt{7},2y+x=2\sqrt{11}+6\sqrt{7}\\ \end{array}$

Centre of the circle is point of intersection of normals$C\left(\frac{8\sqrt{7}}{3},\sqrt{11}+\frac{5\sqrt{7}}{3}\right)$

$\text{ Tangent is }\sqrt{11}y-3x=\frac{5\sqrt{77}}{3}+11$

$$\begin{gathered} \text{ Radius will be }\perp \text{ distance of tangent from center } \\ r=4\sqrt{\frac{7}{5}}\Rightarrow (5h-8k{)}^2+5r^2=816 \end{gathered}$$