Selection tests for the Balkan Mathematical Olympiad 2013

Notice that, if $(x,y)$ is a solution of the inequality, then $(-x,y)$, $(x,-y)$, and $(-x,-y)$ are all solutions of the inequality. Therefore, we can assume $x,y\ge 0$ and deduce the other solutions by symmetries with respect to $x$-axis and $y$-axis.

Assume $x,y\ge 0$. The inequality is equivalent to $$(x-2{)}^2+(y-2{)}^2\le 8.$$ The points whose coordinates satisfy this inequality are precisely the points in the intersection of the first quadrant with the disk of center $(2,2)$ and radius $\sqrt{8}$. By applying the above symmetries we obtain the following surface of points whose coordinates satisfy the inequality.



Its area, is the area of a square of side length $4\sqrt{2}$ and four half disks of radius $2\sqrt{2}$, that is $32+16\pi$.