Question

If radius of the circumcircle of the triangle formed by the lines $x^2-y^2=0$ and $2x-3y=5$ is $a$ units, then value of $a^2$ is

Answer 1

Given pair of lines
$x^2-y^2=0$
$\Rightarrow (x-y)(x+y)=0$
$\Rightarrow x-y=0\cdots \left(1\right)$
and $x+y=0\cdots \left(2\right)$
Also, $2x-3y=5\cdots \left(3\right)$
Solving $\left(1\right),\left(2\right)$ and $\left(3\right),$ we get the vertices,
$P(-5,-5),Q(1,-1),O(0,0)$


Clearly, $OP\perp OQ$
So, $OPQ$ is right angled triangle with $PQ$ as the hypotenuse.
$PQ$ will be the diameter of the circumcircle.
$PQ=\sqrt{(-5-1{)}^2+(-5+1{)}^2}=\sqrt{36+16}=2\sqrt{13}$
Radius $=\frac{PQ}{2}=\frac{2\sqrt{13}}{2}=\sqrt{13}=a$
$\therefore a^2=13$