<s>[INST] 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 !!
 [/INST] By C-S $$x^2+y^2=2(x-y)+2\leq2\sqrt{(1^2+(-1)^2)(x^2+y^2)}+2,$$ which gives
$$x^2+y^2-2\sqrt2\sqrt{x^2+y^2}+2\leq4$$ or
$$\left(\sqrt{x^2+y^2}-\sqrt2\right)^2\leq4,$$ which gives
$$\sqrt{x^2+y^2}\leq2+\sqrt2.$$
Id est, $$x^2+y^2+\sqrt{32}\leq(2+\sqrt2)^2+4\sqrt2=6+8\sqrt2.$$ The equality occurs for $x=1+\sqrt2$ and $y=-1-\sqrt2,$ which says that we got a maximal value.
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