4. On the plane $(x ; y)$, plot the set of points whose coordinates satisfy the system $\left\{\begin{array}{l}(x-|x|)^{2}+(y-|y|)^{2} \leq 4, \\ y+2 x \leq 0\end{array}\right.$ and find the area of the resulting figure.
Answer: $\frac{5+\pi}{4}$.

Solution. Consider the first inequality. There are four possible cases.

1) $x \geq 0, y \geq 0$ (first quadrant). Then $0 \leq 4$, the inequality is satisfied by all points in the first quadrant.
2) $x<0, y \geq 0$ (second quadrant). Then $4 x^{2} \leq 4,|x| \leq 1$; in this case $x$ is negative, so $x \geq-1$.
3) $x<0, y<0$ (third quadrant). Then $4 x^{2}+4 y^{2} \leq 4, x^{2}+y^{2} \leq 1^{2}$. We get the points lying on the circle centered at $O(0 ; 0)$ with radius 1 or inside it.
4) $x \geq 0, y<0$ (fourth quadrant). Then $4 y^{2} \leq 4$; in this case $y$ is negative, so $y \geq-1$.

The second inequality defines the set of points located on the line $y=-2 x$ or below it.

The set of points satisfying the system of inequalities is the union of

- a quarter of a circle centered at $O$ with radius 2, lying in the third quadrant (area $\frac{\pi}{4}$),
- a triangle with vertices $O, A(-1 ; 2), B(-1 ; 0)$ (area 1),
- a triangle with vertices $O, C\left(\frac{1}{2} ;-1\right), D(0 ;-1)$ (area $\left.\frac{1}{4}\right)$.

Thus, the area of the set is $\frac{5+\pi}{4}$.