If real $x$ and $y$ satisfy $x^2+y^2-4x+10y+20=0$, then prove that $y+7-3\sqrt{2}\le x\le y+7+3\sqrt{2}$ 
Let $x, y \in \mathbb{R}$ such that $x^2 + y^2 - 4x + 10y + 20 = 0$. Prove that 
  $$y + 7 - 3\sqrt{2} \le x \le y + 7 + 3\sqrt{2}$$

I'm struggling with that problem. I've recognized that the given equation is one of a circle, namely; $$(x-2)^2 + (y+5)^2 = 9$$
And what needs to be proven can be derived to the form: $$7-3\sqrt{2}\le x-y\le 7 + 3\sqrt{2} \quad\Leftrightarrow\quad (x-y)^2 - 14(x-y) + 31 \le 0$$
But I really can't find the correlation between what's given and what has to be proven! Also, I'm wondering if this problem can be generalized for any circle parameters, something like an universal interval of $x-y$ for any circle. Any help is apprecatiated! Thank you in advance!

Hint:
Let $x-y=c$
Use $x=y+c$   to form a Quadratic Equation  in $y$ 
As $y$ is real, the discriminant must be $\ge0$ 
One may replace $y$ with $x-c$ as well. 
