Shortest separation

We want the point $P$ on the parabola and the point $C$ on the circle so that $CP$ is perpendicular to the tangent to the parabola at $P$ , and perpendicular to the tangent to the circle at $C$ . But then $CP$ is parallel to the radius of the circle at $C$ , and hence passes through the centre $(0,-6)$ of the circle. Thus, if $P$ has coordinates $(x,y)$ , then $CP$ has gradient $\tfrac{y+6}{x}$ . The gradient of the tangent at $P$ is $\tfrac{4}{y}$ , and so we require $\tfrac{y+6}{x} = -\tfrac{y}{4}$ , so that $x = -4 - \tfrac{24}{y}$ . But then $\begin{aligned} y^2 & = \; 8x \; = \; -32 - \tfrac{192}{y} \\ y^3 + 32y + 192 & = \; 0 \end{aligned}$ and the only real solution of this equation is $y=-4$ . Thus $P$ has coodinates $(2,-4)$ , so the distance from $P$ to the centre of the circle is $2\sqrt{2}$ , and hence the minimum distance $CP = \boxed{2\sqrt{2}-1}$ .

Note that $x^2 + (y+6)^2 = 1$ is a circle centered at $(0,-6)$ with radius $1$ . The shortest distance between the circle and parabola is from the negative half of the parabola that is from a point $(x, -\sqrt {8x})$ . Let the distance between the center of the circle and the parabola be $l = \sqrt{(x-0)^2 + (-\sqrt{8x}+6)^2}$ . Then the distance between the circle and parabola is $l - 1$ . And we have:

$\begin{aligned} l - 1 & = {\color{#3D99F6}\sqrt{x^2 + (6-\sqrt{8x})^2}} - 1 & \small \color{#3D99F6} \text{By Titu's lemma} \\ & \ge {\color{#3D99F6} \frac {x + 6-\sqrt{8x}}{\sqrt 2}} - 1 & \small \color{#3D99F6} \text{Equality occurs when }x = 6 - \sqrt{8x} \\ & = {\color{#3D99F6} \frac {2 + 6-\sqrt{16}}{\sqrt 2}} - 1 & \small \color{#3D99F6} \implies x = 2 \\ & = 2\sqrt 2 - 1 = \boxed{1.828} \end{aligned}$

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Reference: Titu's lemma