Let the lines y + 2 x = 11 + 7 7 and 2 y + x = 2 11 + 6 7 be normal to a circle C : x - h 2 + y - k 2 = r 2 …

Equations of normal are

$y+2x=\sqrt{11}+7\sqrt{7} ...\left(i\right)$

$2y+x=2\sqrt{11}+6\sqrt{7} ...\left(ii\right)$

Now the center of the circle is point of intersection of the normals i.e. solving $\left(i\right) \& \left(ii\right)$, we get the point of intersection as 

$\left(\frac{8\sqrt{7}}{3},\sqrt{11}+\frac{5\sqrt{7}}{3}\right)\equiv \left(h,k\right)$

The equation of tangent is $\sqrt{11}y-3x=\frac{5\sqrt{77}}{3}+11$

The radius will be perpendicular distance of tangent from center

i.e. $r=\frac{\left|\sqrt{11}{\displaystyle \frac{8\sqrt{7}}{3}}-3\left(\sqrt{11}+{\displaystyle \frac{5\sqrt{7}}{3}}\right)-{\displaystyle \frac{5\sqrt{77}}{3}}-11\right|}{\sqrt{11+9}}=4\sqrt[]{\frac{7}{5}}$

Hence ${\left(5h-8k\right)}^2+5r^2=816$