Problem
Using Calculus, find the point on the circle $(x-3)^2 + (y-1)^2 = 16$ that is closest to arbitrary point $(-2,2)$ in the $x-y$ plane that is not on the circle.
My Attempt : Attached images show the derivation from which I took the first derivative. I took the derivative of what I believed to be the distance formula from the point to a point on the circle. Please advise.
On
You want to find the point (x,y) on the circle that minimizes
$$D(x) = \sqrt{(x+2)^2+(y+2)^2} \tag{1.}$$
where $$(x-3)^2 + (y-1)^2 = 16\tag{2.}$$
Taking the (implicit) derivative of $(2.)$, we find
\begin{align} 2(x-3) + 2(y-1) y' &= 0 \\ y' &= -\dfrac{x-3}{y-1} \end{align}
We need to solve
\begin{align} D'(x) &= 0 \\ \dfrac{2(x+2)+2(y+2)y'}{2\sqrt{(x+2)^2+(y+2)^2}} &= 0 \\ (x+2)+(y+2)y' &= 0 \\ y' &= -\dfrac{x+2}{y+2} \\ \dfrac{x-3}{y-1} &= \dfrac{x+2}{y+2} \\ xy-3y+2x-6 &= xy-x+2y-2 \\ 3x-5y &= 4 \\ y &= \frac 35x - \frac 45 \end{align}
Substituting this into $(2)$, we get
\begin{align} (x-3)^2 + \bigg(\frac 35x - \frac 95 \bigg)^2 &= 16 \\ (x-3)^2 + \dfrac{9}{25}(x-3)^2 &= 16 \\ (x-3)^2 &= \dfrac{200}{17} \\ x &= 3 \pm 10\sqrt{\dfrac{2}{17}} \\ y &= 1 \pm 6\sqrt{\dfrac{2}{17}} \\ (x,y) &\in \left\{ \bigg(3 + 10\sqrt{\dfrac{2}{17}}, 1 + 6\sqrt{\dfrac{2}{17}} \bigg), \bigg(3 - 10\sqrt{\dfrac{2}{17}}, 1 - 6\sqrt{\dfrac{2}{17}} \bigg) \right\} \end{align}
One solution gives you the coordinates of the closest point and the other gives you the coordinates of the furthest point.
The line that passes through $(—2,—2)$ and the center of the circle intersects the circle at the closest and the farthest point.