How to draw a set $\{(x,y): y^2 \leq x^2\}$

Question

I want to draw a set $\{(x,y): y^2 \leq x^2\}$. I know what the result should look like (the blue region):

But I don't really see why is that so. I have

$$ y^2 \leq x^2 $$

I am dividing the domain in two: for $x<0$ and $x \geq 0$ and after square root I get: $$\pm y \leq \pm x$$ that is: $$y \leq x$$ $$-y \leq x => y \geq -x$$

and that is fine, but I also get:

$$y \leq -x$$

which is below line for $x$ greater than zero. What am I doing wrong?

Answer 1

Since $\sqrt{}$ is an increasing function, $y^2 < x^2$ is equivalent to $\sqrt{y^2} < \sqrt{x^2}$, i.e. $|y| < |x|$. Since $|y| = \max(y, -y)$, that is equivalent to ($y < |x|$ and $-y < |x|$). But $-y < |x|$ is equivalent to $y > -|x|$, so the condition is $|x| > y > -|x|$. That is, for $x \ge 0$, $x > y > -x$, and for $x \le 0$, $-x > y > x$.

Answer 2

$${\bf x,y\geq0:}\\ x\geq y\\ {\bf x,-y\geq0:}\\ x\geq -y\\ {\bf -x,-y\geq0:}\\ x\leq y\\ {\bf -x,y\geq0:}\\ -x\geq y$$