Problem regarding the distance between a circle and a parabola The question is this, copy pasted here:
Consider the circle $C$ whose equation is
$$(x - 2) ^ 2 + (y - 8) ^ 2 = 1$$
and the parabola $P$ with the equation
$$y^ 2 =4x.$$
I have no idea how to model the question, I tried showing that the $x-$coordinates of the two points have to be the same but that seems to be wrong too? Can anyone help?
 A: Some hints:

*

*Your first move should indeed be to draw the circle and parabola, as in the answer by Jose Arnaldo Bebita Dris (which seems to be deleted now). You will be able to estimate the region of closest approach and the minimum distance. I have sketched it here:

The green line is a reasonable guess, and its length is around 3.5. So this will give you an intuition what the answer should be.


*You're looking for a straight line connecting a point on the circle and a point on the parabola (both of which are smooth curves). Can you convince yourself that the line has to intersect both curves at right angles? (Ask yourself: If it's not a right angle, can you find a shorter line?)
b. As a corollary, the slopes of the circle and parabola at theri repsective intersection with the line need to be equal, since they're both orthogonal to a straight line. So the idea in your comment is correct.


*Can you convince yourself that, as a consequence of the previous point, that the line has to go through the centre of the circle? And that you can thus also find the shortest path from the centre to the parabola?


*If you have thusly convinced yourself, can you express all lines through the centre of the circle by some angle $\alpha$ and find the intersection to the intersection with the parabola.
You could now either calculate the distance and minimise that (it's a function of one variable only) or check whether the line intersects the parabola at a right angle.
Can you take it from here?
A: Geometric Analysis:

As we approach from external points to a circle, the distance will be less and less the more we go to the center of the circle.

Geometry and Calculus:
Consider the parametric form of the parabola $$P:\quad y^{2}=4x$$ which is given by $$\gamma(t)=(t^{2},2t)$$
The circle $$C:\quad (x-2)^{2}+(y-8)^{2}=1$$
with center $O(2,8)$ and radius $r=1$.
Now
$$d((t^{2},2t),(2,8))=\sqrt{(t^{2}-2)^{2}+(2t-8)^{2}}=\sqrt{t^{4}-32t+68}$$
Then the distance will be the minimum when $t^{4}-32t+68$ is the minimum.
The first derivative equals to $0$ is given by $4(t^{3}-2^{3})=0$ or equivalently $(t-2)(t^{2}+2t+4)=0$ so $t=2$ or $t=-1\pm i\sqrt{3}$. We are left with the real root $t=2$ and notice that second derivative at $t=2$ give $12(2)^{2}>0$ so by the second derivative criterion $t=2$ is a minimum.
Hence with $t=2$ we have the point $(2^{2},2\cdot 2)$ i.e., the point $\boxed{(4,4)}$. Therefore the distance between $P$ and $C$ is given by $d=\sqrt{(4-2)^{2}+(4-8)^{2}}-r=\boxed{2\sqrt{5}-1}$.
That's not the unique approach, also we can use the following fact geometric
Geometric Analysis:

The minimum distance between $P$ and $C$ is the minimum length of the line segment in the common normal that $P$ and $C$ will share.

Geometry and Calculus:
The equation of the normal for the parabola $(t^{2},2t)$ will be
$$y-2t=-t(x-t^{2})$$
Now all normal of a circle passes through its center. So in our case the equation of common normal should pass through the center of the circle., that is, satisfies
$$8-2t=-t(2-t^{2})\implies t^{3}-2^{3}=0\implies t=2,\quad t=-1\pm i\sqrt{3}$$
We are left with the real root and get the point $(4,4)$ so should be again $d=2\sqrt{5}-1$, you can complete the details.
A: Differentiating $y^2=4x$ gives us $\dfrac{dy}{dx}=\dfrac{2}{y}$. And the gradient of the line between $(2,8)$ and $(x,y)$ is $\dfrac{y-8}{x-2}$; subsituting $x=y^2/4$, this is $\dfrac{4y-32}{y^2-8}$.
We want the gradient of this line to be perpendicular to the slope of the parabola at $(x,y)$, so
$$\frac{2}{y}\cdot\frac{4y-32}{y^2-8}=-1$$
This is a cubic equation in $y$, which simplifies to $y^3=64$, i.e. $y=4$. So the point of closest approach to the circle is $(x,y)=(4,4)$. You can do the rest.
A: Let $(x,y) $ be on the parbola, then
$ y^2 = 4 x$
The normal vector is $[ 4, -2 y] $ , while the vector from $(2, 8)$ to $(x,y)$ is $[ x - 2, y - 8 ]$, and we want these two vectors to be aligned; so we can use the determinant and set it equal to zero, namely,
$4 (y-8) + 2 y (x - 2) = 0 $
We have to solve the above equation together with $ y^2 = 4 x $
Hence,
$ 4 ( y - 8) + 2 y (y^2/4 - 2) = 0 $
From which
$ -64 + y^3 = 0$
Whose solution is $ y = 4 $, and therefore, $ x = y^2 /4 = 4 $
This means that the shortest distance between the circle and the parabola is
$ d = \sqrt{ (2 - 4)^2 + (8 - 4)^2 } - 1 = \sqrt{20} - 1 = 2 \sqrt{5} - 1 $
A: Part 1. Show that shortest distance line passes through center of circle (separate investigation not used in part 2)
Let $A$ is point of parabola which is end of shortest distance between curves and $B$ is point of circle which is another end of shortest distance. Suppose $B$ is not between $A$ and $C$ (center of circle). Consider then point $B_1$ which is intersection of circle with edge $AC$. One can easily show that $AB_1 < AB$. Then $AB$ is not shortest distance between curves. Contradiction. Then $B$ must be between $A$ and $C$ then line $AB$ must pass through $C$.
Part 2. Direct calculation from definition of distance between curves.
Arbitrary point on circle: $$x_c=2+\cos t, y_c=8+\sin t$$
Arbitrary point on parabola: $$x_p=u^2, y_p=2u$$
Square of distance between points: $$d^2=(x_p-x_c)^2+(y_p-y_c)^2=(u^2-2-\cos t)^2+(2u-8-\sin t)^2=\\
u^4-32u+69-(2u^2-4)\cos t-(4u-16)\sin t$$
$$(a\cos t+b\sin t)^2+(a\sin t-b\cos t)^2=a^2+b^2 \Rightarrow a\cos t+b\sin t \leq a^2+b^2$$
Equality is reached at $a\sin t-b\cos t=0$.
$$d^2=u^4-32u+69-(2u^2-4)\cos t-(4u-16)\sin t\geq \\
u^4-32u+69-\sqrt{(2u^2-4)^2+(4u-16)^2}=u^4-32u+69-2\sqrt{u^4-32u+68}$$
Let $\sqrt{u^4-32u+68}=n$, then $$d^2\geq n^2+1-2n=(n-1)^2 \Rightarrow d\geq |n-1|$$
$$u^4-32u+68=u^4-8u^2+16+8u^2-32u+32+20=(u^2-4)^2+8(u-2)^2+20\geq 20$$
$$n\geq \sqrt{20} \Rightarrow d\geq \sqrt{20}-1$$
Equality is reached at $u=2$ and $(2u^2-4)\cos t+(4u-16)\sin t=\sqrt{(2u^2-4)^2+(4u-16)^2}$, $t=-\arctan 2$
