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

Correct Answer: 816

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 \(816\).