Given that $x^2 + y^2 = 2x - 2y + 2$ , find the maximum value of $x^2 + y^2 + \sqrt{32}$ . 
Given that $x^2 + y^2 = 2x - 2y + 2$ , find the maximum value of $x^2 + y^2 + \sqrt{32}$ .


What I Tried :- Since $x^2 + y^2 = 2x - 2y + 2$ , we have $2x - 2y + 2 + \sqrt{32}$
=> $2(x - y + 1 + 2√2)$ . From this step I am not sure how to move forward .
Also I tried to express $x^2 + y^2 + \sqrt{32} \leq S$ , so that in that way we can say that $x^2 + y^2 + \sqrt{32}$ is maximum at $S$ , but I couldn't do it .
Can anyone help me ? Some hints or suggestions to this problem will be appreciated !!

After square completion, there is a simple solution just using the triangle inequality for the Euklidean distance:
Using $(x-1)^2 + (y+1)^2 =4$ we get
$$x^2+y^2= \left(\left|\binom xy\right|\right)^2 \stackrel{\triangle-ineq.}{\leq} \left(\left|\binom{x-1}{y+1}\right| + \left|\binom{-1}{1}\right|\right)^2= (2+\sqrt 2)^2=6+4\sqrt2$$
Now, just add $\sqrt{32}$.
