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
$\sqrt {11}y - 3x =\frac { 5\sqrt {17}}{3} + 11$
is tangent to the circle C , then the value of $(5h – 8k)^2 \+ 5r^2$ is equal to _______.

Answer 1

Line 1 : $y + 2x = \sqrt {11} + 7\sqrt 7$
Line 2 : $2y + x = 2\sqrt {11} + 6\sqrt 7$
Point of intersection of these two lines is centre of circle i.e.
$( \frac 83√7, √11 + \frac 53√7 )$
Perpendicular from centre to line $3x − \sqrt {11}y + (\frac {5\sqrt {77}}{3} + 11 ) = 0$
is radius of circle
$⇒ r =|\frac { 8\sqrt 7 − 11 − \frac 53\sqrt {77} + \frac {5\sqrt {77}}{3} + 11}{\sqrt {20}}|$

$= | \sqrt[4]{\frac 75} | = \sqrt[4]{\frac 75}$ units
So $(5h – 8K)^2 \+ 5r^2$
$=(\frac {40}{3}√7 − 8\sqrt {11} − \frac {40}{3}\sqrt 7 )^2 + 5.16 .\frac 75$
$= 64 × 11 + 112$
$= 816$

So, the answer is $816$.