Find largest possible value of $x^2+y^2$ given that $x^2+y^2=2x-2y+2$ 
Let $x, y \in \mathbb R$ such that $x^2+y^2=2x-2y+2$.  Find the largest possible value of $x^2+y^2$.

My attempt:
$x^2+y^2=2x-2y+2$
$(x^2-2x)+(y^2+1)=2$
$(x-1)^2+(y+1)^2=4$
I have no idea how to continue here.  Any help?

Thanks to StackTD, I have achieved my solution.
$x^2+y^2=2x-2y+2$
$(x^2-2x)+(y^2+1)=2$
$(x-1)^2+(y+1)^2=4$
We need to find the maximum distance from any point on the circle to the origin, squared(to find $x^2+y^2$).
By Pythagoras' theorem, the distance from the centre of the circle to the origin is $\sqrt{(-1)^2+1^2}=\sqrt2$.  We then add $2$, the radius of the circle.  We have $2+\sqrt2$, the maximum distance on any point on the circle to the origin.   However, this is only $\sqrt{x^2+y^2}$, so we need to square $2+\sqrt2$, as mentioned above, and will get $6+4\sqrt2$ as answer.
