If the shortest distance of the parabola y ^ 2 = 4x from the centre...

Given: Equation of parabola is ${y}^{2} = 4x$  and the equation of circle is ${x}^{2} + {y}^{2} - 4x - 16y + 64 = 0$.

We know that, equation of normal to parabola is given by, $y = mx - 2m - {m}^{3}$.

Also, the centre of given circle is \left(2, 8\right)$.</p><p>The normal of parabola is passing through center of circle.</p><p>$⇒ 8 = 2m - 2m - {m}^{3}$</p><p>$⇒ m = - 2$</p><p>So point$P$on parabola is given by,</p><p>\left({am}^{2}, - 2am\right) = ( 4, 4 )$

And $C = ( 2, 8 )$

$⇒ PC = \sqrt{4 + 16} = \sqrt{20}$

${d}^{2} = 20$.