Let $x, y \in \mathbb R$ such that $x^2+y^2=2x-2y+2$. Find the largest possible value of $x^2+y^2$ 
Let $x, y \in \mathbb R$ s. t . $x^2+y^2=2x-2y+2$.  Find the largest possible value of $x^2+y^2-\sqrt{32}$

I know this is a duplicate of another question, but that question has solutions involving calculus and geometry, while I want a solution relying on algebra and basic inequalities only to solve this problem.
 A: You can rearrange the equation into the form
$$(x-1)^2+(y+1)^2=r^2$$
for some constant $r$ which I will let you find for yourself. This is a circle with centre $(1,-1)$ and radius $r$, so obviously its furthest distance from the origin is $r+\sqrt 2$.
A: Let $ z=x+iy$. Then, we have $ |z-(1+i)| =2$ from $x^2+y^2=2x-2y+2$, and
$$|z|= |z-(1+i)+(1+i )| \le | z-(1+i)|+ |1+i |= 2+ \sqrt 2=|z|_{max}$$
Thus, the largest possible value is
$$x^2+y^2= |z|_{max}^2 = (2+\sqrt 2)^2=6+4\sqrt2$$
A: We have: $$6+ \underbrace{\left( 1+\frac{\,\sqrt {2}}{2} \right)  \left( {x}^{2}+{y}^{2}-2\,x+2\,y-2
 \right)}_{=\, 0} -({x}^{2}+{y}^{2}-\sqrt {32})$$
$$=\frac{\,\sqrt {2}}{4} \left( \sqrt {2}\,x-\sqrt {2}-2 \right) ^{2}+\frac{\sqrt{2}}{4}\,
 \left( \sqrt {2}y+\sqrt {2}+2 \right) ^{2}\geqq 0$$
Therefore: $$x^2 +y^2 -\sqrt{32} \leqq 6$$

Use this useful identity: $$ax^2 +bx+c =a(x+\frac{b}{2a})^2 +\frac{4ac-b^2}{4a}$$

A: Yet another approach: Let $S=\{(x,y) : x^2+y^2=2x-2y+2\}$ be the algebraic curve of the constraint and let $A=\{(x,y): x+y=0\}$ be the anti-diagonal. We have the basic inequality: $(x-y)^2 \leq 2(x^2+y^2)$ with equality iff $(x,y)\in A$. Thus, for points in $S$ we have:
  $$ (x-y)^2 \leq 4(x-y) + 4 $$
with equality iff $(x,y)\in A$ (in addition to being in $S$).
Equivalently,  $(x-y-2)^2 \leq 8$ or $2-2\sqrt{2} \leq x-y \leq 2+2\sqrt{2}$, with the two extremal cases corresponding to $S\cap A$. Evaluating on $S$ we get the upper bound:
  $$ x^2+y^2 -\sqrt{32} \leq 2(2+2\sqrt{2}) + 2 - 4\sqrt{2} = 6$$
with equality  being attained at the unique intersection of $x-y=2+2\sqrt{2}$ and $x+y=0$, i.e. $x=-y=1+\sqrt{2}$
