2. Find all values of $a$, for each of which the system

$$
\left\{\begin{array}{l}
(x-2)^{2}+(|y-1|-1)^{2}=4 \\
y=b|x-1|+a
\end{array}\right.
$$

has solutions for any value of $b$
# Problem 2.

Answer: $-\sqrt{3} \leqslant a \leqslant 2+\sqrt{3}$.

Solution. Let's represent the solution of the system on the coordinate plane. The first equation of the system defines the union of two arcs of circles: $(x+1)^{2}+(y-2)^{2}=4, y \geqslant 1$ and $(x+1)^{2}+y^{2}=4, y<1$. The second equation of the system, when $b=0$, defines the line $y=a$ on the plane, and when $b \neq 0$, it defines two rays $y=b(x+2)+a, x \geqslant-2$ and $y=-b(x+2)+a, x<-2$ with a common origin at the point $(-2, a)$.

The line $x=-2$ intersects the arcs of the circles at points $A(-2,2+\sqrt{3})$ and $B(-2,-\sqrt{3})$. Therefore, for the system to have a solution, it is necessary and sufficient that the common origin of the rays lies on the segment $A B$.