1. Determine the area of the figure that is defined in the Cartesian coordinate system by the inequalities:

$$
\begin{aligned}
x^{2}+y^{2} & \leqslant 4(x+y-1) \\
y & \leqslant \sqrt{x^{2}-4 x+4}
\end{aligned}
$$
1. The first inequality reduces to $x^{2}-4 x+$ $4+y^{2}-4 y \leqslant 0$, i.e., $(x-2)^{2}+(y-2)^{2} \leqslant 4=2^{2}$. The set of points that satisfy this represents a circle with center at $(2,2)$ and radius 2. The second inequality reduces to $y \leqslant$ $\sqrt{(x-2)^{2}}=|x-2| ;$ thus, the figure in question represents a part of the circle described earlier that lies below the graph of the function $y=|x-2|$, i.e., it consists of two circular segments (see the figure). If we mark the points $A(0,2)$, $B(2,0)$, and $C(4,2)$, we can easily see that the desired area can be calculated by subtracting the area of the isosceles right-angled $\triangle A B C$ (with one leg equal to $2 \sqrt{2}$) from half the area of the circle, i.e., it is $\frac{\pi \cdot 2^{2}}{2}-\frac{(2 \sqrt{2})^{2}}{2}=2 \pi-4$.

![](https://cdn.mathpix.com/cropped/2024_06_04_841f52b4ffb037ad799bg-57.jpg?height=500&width=507&top_left_y=690&top_left_x=1011)

Ok 2018 4B 1