If the shortest distance of the parabola $y^2=4x$ from the centre of the circle $x^2+y^2-4x-16y+64=0$  is d, then d² is equal to :

Equation of normal to parabola

$y=mx-2m-m^3$

this normal passing through center of circle (2, 8)

$\begin{array}{rl} & 8=2\mathrm{m}-2\mathrm{m}-{\mathrm{m}}^3\\ & \mathrm{m}=-2\end{array}$

So point P on parabola $\Rightarrow ({\mathrm{am}}^2,-2\mathrm{am})=(4,4)$

And C = (2, 8)

$\mathrm{PC}=\sqrt{4+16}=\sqrt{20}$

${\mathrm{d}}^2=20$