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

Let the lines y 2x 11 7 7 and 2y x 2 11 6 7 be normal to a c: (Mathematics)

${L_1}:y + 2x = \sqrt {11} + 7\sqrt 7$

${L_2}:2y + x = 2\sqrt {11} + 6\sqrt 7$

Point of intersection of these two lines is centre of circle i.e. $\\ \left( {{8 \over 3}\sqrt 7 ,\sqrt {11} + {5 \over 3}\sqrt 7 } \right)$

$\\ { \bot ^r}$ from centre to line $3x - \sqrt {11} y + \left( {{{5\sqrt {77} } \over 3} + 11} \right) \\= 0$ is radius of circle

$\Rightarrow r = \left| {{{8\sqrt 7 - 11 - {5 \over 3}\sqrt {77} + {{5\sqrt {77} } \over 3} + 11} \over {\sqrt {20} }}} \right|$

$= \left| {root 4 of {{7 \over 5}} } \right| =root 4 of {{7 \over 5}}$ units

So ${(5h - 8K)^2} + 5{r^2}$

$= {\left( {{{40} \over 3}\sqrt 7 - 8\sqrt {11} - {{40} \over 3}\sqrt 7 } \right)^2} + 5.\,16.\,{7 \over 5}$

$= 64 \times 11 + 112 = 816$ .