Find minimum value $E(x, y) = x^2 + y^2 -6x -10y$, where $x^2 + y^2 - 2y \le 0$ I am given the expression:
$$E(x, y) = x^2 + y^2 -6x -10y$$
And I have to find the minimum value of $E(x, y)$ for $(x, y) \in D$ where:
$$D = \{ (x,y) \in \mathbb{R}^2 \hspace{0.25cm} | \hspace{0.25cm} x^2 + y^2 -2y \le 0 \}$$
Doing the following:
$$\hspace{5.8cm} x^2+y^2-2y \le 0 \hspace{5cm} |+1$$
$$x^2+(y-1)^2 \le 1$$
So I have to find the minimum value of $E(x,y)$ where $(x, y)$ is from the circle $x^2 + (y-1)^2 \le 1$.
I don't see how I should approach this.
 A: since the other is $(x-3)^2 + ( y-5)^2 + \mbox{constant},$ they are asking for the smallest(squared) distance from $(3,5)$ on the disk $x^2 + (y-1)^2 \leq 1.$
Draw some pictures, just circles and line segments are enough. In particular, draw the line that joins the two circle centers.   

A: Note that $E(r) = r^2-34$, where $r$, given by the circle 
$r^2= (x-3)^2 + (y-5)^2$, is its radius and the circle $x^2+(y-1)^2=1$ has radius 1.
So, $E(r)$ is minimum if $r$ is the smallest, which is the case when the two circles just touch, i.e. the sum of their radii is equal to the distance of their centers, $r+1=\sqrt{3^2+4^2}=5$. Thus, the minimum $E$ is $(5-1)^2-34$.
A: Hint.
The following plot shows in black the level curves for $x^2+y^2-6x-10y$ and in light blue the feasible region $x^2+(y-1)^2 \le 1$ The solution is found at the tangency points.

A: For $(x,y)=\left(\frac{3}{5},\frac{9}{5}\right)$ we'll get a value $-18$.
We'll prove that it's a minimal value.
Indeed, it's enough to prove that:
$$x^2+y^2-6x-10y\geq-18+4(2y-x^2-y^2)$$ or
$$5x^2-6x+5y^2-18y+18\geq0$$ or
$$5\left(x-\frac{3}{5}\right)^2+5\left(y-\frac{9}{5}\right)^2\geq0$$ and we are done!
