Why is $ y \le 0$ one of the solutions of the equation $\sqrt{x^2+y^2} = -y $? Or is the answer (at the end of my book) wrong. I know that the other solution is x=0, but $y \le 0$ can't be verified.
After some calculations we arrive at $ y = -y$... but if we substitute with $-3$, for example, we get: $ -3 = -(-3) \rightarrow -3=3 $.
A solution of this equation is an ordered pair $(x,y)$. Now observe that squaring both sides gives $x=0$. So you are left with the equation $\sqrt{y^2}=-y$. Now, do you know that one of the definitions of absolute value is $|y|=\sqrt{y^2}$? Combine this with the piecewise function definition of $|y|$, and you will get $y\leq 0$.