Explanation

(c)
Let $(x,\ y)$ be any point on the parabola $2y = x^{2}$
Let $f(x) = (x - 0)^{2} + (y - 3)^{2} = x^{2} + \left( \frac{x^{2}}{2} - 3 \right)^{2}$
$\Rightarrow \ \ \ \ \ \ f^{'}(x) = 2x + 2\left( \frac{x^{2}}{2} - 3 \right)(x)$
Put $f^{'}(x) = 0\ \ \ \Rightarrow \ \ \ 2x\left( 1 + \frac{x^{2}}{2} - 3 \right) = 0$
$\Rightarrow \ \ \ \ \ \ \ x = 0,\ \pm 2$
Now, $f^{''}(x) = 2 + 2\left( \frac{x^{2}}{2} - 3 \right) + 2x(x)$
At $x = 0,\ \ \ \ \ f^{''}(x) = 2 - 6 < 0$, maximum
At $x = 2,\ \ \ \ f^{''}(x) > 0$, minimum
At $x = - 2,\ \ \ f^{''}(x) > 0$, minimum
$\therefore$ At $x = \pm 2\ \ \ \Rightarrow \ \ \ \ y = 2$
$\therefore$ Required point is $( \pm 2,\ 2)$