1. Compute the area of the figure defined by the inequality

$$
x^{2}+y^{2} \leqslant 4(x+y)
$$
Answer: $8 \pi$.

Solution. If we rewrite the inequality as

$$
(x-2)^{2}+(y-2)^{2} \leqslant 8
$$

it is easy to see that it defines a circle, the square of the radius of which is 8.

Evaluation. 10 points for the correct solution.