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

Question

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.

Answer 1

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 $r+\sqrt 2$.

Answer 2

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

$$x^2+y^2= |z|_{max}^2 = (2+\sqrt 2)^2=6+4\sqrt2$$

Answer 3

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

Answer 4

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. $x=-y=1+\sqrt{2}$