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 !!

Hint: $x^2+y^2=2x-2y+2$ is equivalent to $(x-1)^2+(y+1)^2=4$, which is the equation of a circle with the centre in $C=(1,-1)$ and radius $2$. To maximise $x^2+y^2+\sqrt{32}$ it is enough to first maximise $\sqrt{x^2+y^2}$, i.e. the distance from the origin $O=(0,0)$.
Now, which point in that circle is the farthest from the origin? It will be the intersection of half-line $[OC)$ with the circle, i.e. the point $(x,y)=(1+\sqrt{2},-1-\sqrt{2})$, giving you the maximum value $x^2+y^2+\sqrt{32}=6+8\sqrt{2}$.
