Let the lines y + 2x = 11 + 7 7 and 2y + x = 2 11 + 6 7 be normal...

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 $\left(i\right) \text{ and } \left(ii\right)$, we get the point of intersection as

\left(\dfrac{8\sqrt{7}}{3}, \sqrt{11} + \dfrac{5\sqrt{7}}{3}\right) ≡ \left(h, k\right)$</p><p>The equation of tangent is$\sqrt{11}y - 3x = \dfrac{5\sqrt{77}}{3} + 11$</p><p>The radius will be perpendicular distance of tangent from center</p><p>i.e.$r = \dfrac{\left|\sqrt{11}\dfrac{8\sqrt{7}}{3} - 3\left(\sqrt{11} + \dfrac{5\sqrt{7}}{3}\right) - \dfrac{5\sqrt{77}}{3} - 11\right|}{\sqrt{11 + 9}} = 4 \dfrac{7}{5}$</p><p>Hence${\left(5h - 8k\right)}^{2} + 5{r}^{2} = 816