A circle of radius 2 unit passes through the vertex and the focus...

The vertex and focus of parabola ${y}^{2} = 2x$ are V\left(0, 0\right)$ andS\left(\dfrac{1}{2}, 0\right)$respectively</p><p>Let the equation of circle passing through vertex and focus of the parabola be${\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = 4$</p><p>$∵$Circle passes through \left(0, 0\right)$

$⇒ {h}^{2} + {k}^{2} = 4 ⋯$\left(1\right)$</p><p>$∵$Circle passes through \left(\dfrac{1}{2}, 0\right)$

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

$⇒ {h}^{2} + {k}^{2} = h + \dfrac{15}{4} ⋯ \left(2\right)$

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

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

$⇒ h = \dfrac{1}{4}$

$⇒ k = \dfrac{\sqrt{63}}{4}$

here $k = - \dfrac{\sqrt{63}}{4}$ is rejected as circle with centre \left(\dfrac{1}{4}, - \dfrac{\sqrt{63}}{4}\right)$can't touch the parabola${y}^{2} = 2x$.</p><p>So the equation of circle will be${\left(x - \dfrac{1}{4}\right)}^{2} + {\left(k - \dfrac{\sqrt{63}}{4}\right)}^{2} = 4$</p><p>Since this circle touches the parabola$y = {\left(x - \dfrac{1}{4}\right)}^{2} + \alpha$</p><p>i.e.$\alpha = 2 + \dfrac{\sqrt{63}}{4} = \dfrac{8 + \sqrt{63}}{4}$</p><p>$4\alpha - 8 = \sqrt{63}$</p><p>${\left(4\alpha - 8\right)}^{2} = 63$