A circle of radius 2 unit passes through the vertex and the focus of the parabola y 2 = 2 x and touches the…

The vertex and focus of parabola $y^2=2x$ are $V\left(0,0\right)$ and $S\left(\frac{1}{2},0\right)$ respectively

Let the equation of circle passing through vertex and focus of the parabola be ${\left(x-h\right)}^2+{\left(y-k\right)}^2=4$

$\because$ Circle passes through $\left(0,0\right)$

$\Rightarrow h^2+k^2=4 \cdots \left(1\right)$

$\because$ Circle passes through $\left(\frac{1}{2},0\right)$

${\left(\frac{1}{2}-h\right)}^2+k^2=4$

$\Rightarrow h^2+k^2=h+\frac{15}{4} \cdots \left(2\right)$

On solving $\left(1\right)$ and $\left(2\right)$

$4=h+\frac{15}{4}$

$\Rightarrow h=\frac{1}{4}$

$\Rightarrow k=\frac{\sqrt{63}}{4}$

here $k=-\frac{\sqrt{63}}{4}$ is rejected as circle with centre $\left(\frac{1}{4},-\frac{\sqrt{63}}{4}\right)$ can't touch the parabola $y^2=2x$.

So the equation of circle will be ${\left(x-\frac{1}{4}\right)}^2+{\left(k-\frac{\sqrt{63}}{4}\right)}^2=4$

Since this circle touches the parabola $y={\left(x-\frac{1}{4}\right)}^2+\alpha$

i.e. $\alpha =2+\frac{\sqrt{63}}{4}=\frac{8+\sqrt{63}}{4}$

$4\alpha -8=\sqrt{63}$

${\left(4\alpha -8\right)}^2=63$