Question

Let $PQ$ be the focal chord of the parabola $y^2=4x.$ If the center of the circle having $PQ$ as its diameter lies on the line $y=\frac{4}{\sqrt{5}},$ then the radius (in units) is equal to

Answer 1

Answer: 3.6

Let, $P$ is $\left(t^2,2t\right)$ then,
$Q$ is $\left(\frac{1}{t^2},\frac{-2}{t}\right)$
Midpoint of $PQ$ is $\left(\frac{t^2+\frac{1}{t^2}}{2},\frac{2t-\frac{2}{t}}{2}\right)$
$\Rightarrow t-\frac{1}{t}=\frac{4}{\sqrt{5}}\Rightarrow t^2+\frac{1}{t^2}-2=\frac{16}{5}$
$\Rightarrow t^2+\frac{1}{t^2}+2=\frac{36}{5}$
$\Rightarrow \left(t^2+1\right)+\left(\frac{1}{t^2}+1\right)=7.2$
If $S$ is the focus, then
$PS+SQ=7.2\Rightarrow PQ=7.2$
Hence, radius $=3.6$ units