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 !!

Using the Lagrange multiplier method, we make the expression $F(x,y,\lambda)=x^2+y^2+\sqrt{32}+\lambda(x^2+y^2-2x+2y-2)$ and then calculate partial derivatives on $x$, $y$ and $\lambda$, require them to be $0$, then solve for $x$ and $y$ (and $\lambda$):
$$0=\frac{\partial F}{\partial x}=2x+2\lambda x-2\lambda$$
$$0=\frac{\partial F}{\partial y}=2y+2\lambda y+2\lambda$$
$$0=\frac{\partial F}{\partial\lambda}=x^2+y^2-2x+2y-2$$
So, $x=\frac{\lambda}{1+\lambda}$ and $y=-\frac{\lambda}{1+\lambda}$ and so $y=-x$, which you can in turn substitute in the third equation. You will end up with two solutions: $(1+\sqrt 2,-1-\sqrt 2)$ and $(1-\sqrt 2,-1+\sqrt 2)$, our of which the first one gives the larger value of $x^2+y^2+\sqrt{32}$ - I will leave it to you to finish off the calculation.
