A circle of radius 2 unit passes through the vertex and the focus of the parabola ${\mathrm{y}}^2=2\mathrm{x}$ and touches the parabola $y={\left(x-\frac{1}{4}\right)}^2+\alpha$, where $\alpha >0$. Then $(4\alpha -8{)}^2$ is equal to_____

Vertex and focus of parabola $y^2=2x$ are $\mathrm{V} (0, 0)$ and $S\left(\frac{1}{2}, 0\right) resp$.
Let equation of circle be

$(x-h{)}^2+(y-k{)}^2=4$

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

$\Rightarrow {\mathrm{h}}^2+{\mathrm{k}}^2=4 \dots \dots \left(1\right)$

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

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

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

On solving (1) and (2)

$\begin{array}{l}4-h=\frac{15}{4}\\ \\ h=4-\frac{15}{4}=\frac{1}{4}\\ \\ k=+\frac{\sqrt{63}}{4}\end{array}$

$k=-\frac{\sqrt{63}}{4}$ is rejected as circle with centre $\left(\frac{1}{4},-\frac{\sqrt{63}}{4}\right)$ can't touch given parabola. Equation of circle is

${\left(x-\frac{1}{4}\right)}^2+{\left(k-\frac{\sqrt{63}}{4}\right)}^2=4$

From figure

$\begin{array}{l}\alpha =2+\frac{\sqrt{63}}{4}=\frac{8+\sqrt{63}}{4}\\ \\ 4\alpha -8=\sqrt{63}\\ \\ (4\alpha -8{)}^2=63\end{array}$