<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] The given expression can be written as $\ (x-1)^2+(y+1)^2=2^2$
$\ x-1=2cos\theta$, $\ y+1=2sin\theta$
$\ x^2+y^2+4\sqrt2=(2cos\theta +1)^2+(2sin\theta-1)^2+4\sqrt2$
$\ =6+4\sqrt2+4(cos\theta-sin\theta)$
Therefore maximum value$\ =6+8\sqrt2$
</s>