# Solving a quadratic inequality geometrically

I have to plot the set $$M=\left\lbrace(x,y)\in\mathbb R\times\mathbb R:(x+y)^2\leq 2|x|y-4+4x+2y\;\wedge\;\left((x\geq0)\vee (y\leq0)\right)\right\rbrace$$ (in the cartesian plane).

What I've already found out so far is that we can add $-4x+4$ to both sides by which we obtain the new inequality $(x-2)^2+2xy+y^2\leq 2|x|y+2y$. How can I proceed from here? Especially, I don't quite understand how to use the additional requirement of $(x\geq0)\vee (y\leq0)$.

EDIT: I just noticed that adding $-2y+1-2xy$ on both sides creates an inequality which looks very similar to the ones used to describe a circle: $(x-2)^2+(y-1)^2\leq 2|x|y-2xy+1$. Does that help?
The statement $(x\geq0)\vee (y\leq0)$ is referring to the quadrants $I, III$ and $IV$.The only time this statement is false is when $(x,y)$ is in the second quadrant with $x <0$ and $y>0$.
In the $I$st and the $IV$th quadrant $|x|=x$, therefore the given inequality becomes $$(x-2)^2+(y-1)^2 \leq 1$$ This is the interior of the circle centered at $(2,1)$ and radius $1$. But this lies completely in the first quadrant only. Hence for the first quadrant the set will contain all the point on and inside this circle, whereas the boolean statement is false for the $IV$th quadrant.
In the $III$rd quadrant, since $|x|=-x$, therefore the inequality becomes $$x^2+y^2+4xy+4-4x-2y \leq 0.$$ But with $x<0$ and $y < 0$, this will not hold.
Based on this, the only points in $M$ are the points on and inside the circle $(x-2)^2+(y-1)^2 \leq 1$.