Question

Consider the circle $x^2+y^2=9$ and the parabola $y^2=8x$. They intersect at $P$ and $Q$ in the first and fourth quadrants, respectively. Tangents to the circle at $P$ and $Q$ intersect the x-axis at $R$ and tangent to the parabola at $P$ and $Q$ intersect the x-axis at $S$.
The radius of the circumcircle of the triangle $PRS$ is-

Answer 1

Equation of circumcircle of $△PRS$

$(x+1)(x-9)+y^2+\lambda y=0$

It will pass through $(1,2\sqrt{2})$

then,

$(1+1)(1-9)+(2\sqrt{2}{)}^2+\lambda .2\sqrt{2}=0$

$-16+8+\lambda .2\sqrt{2}=0$

$\lambda =\frac{8}{2\sqrt{2}}=2\sqrt{2}$

Equation of circumcircle is

$x^2+y^2-8x+2\sqrt{2}y-9=0$

Hence radius is $3\sqrt{3}$