7. Plot the figure $\Phi$ on the plane, consisting of points $(x ; y)$ of the coordinate plane such that the system of inequalities is satisfied

$$
\left\{\begin{array}{l}
\sqrt{x^{2}-3 y^{2}+4 x+4} \leqslant 2 x+1 \\
x^{2}+y^{2} \leqslant 4
\end{array}\right.
$$

Determine how many parts the figure $\Phi$ consists of.
Solution. The first inequality is equivalent to the system of inequalities

$$
\left\{\begin{array} { l } 
{ x ^ { 2 } - 3 y ^ { 2 } + 4 x + 4 \leqslant 4 x ^ { 2 } + 4 x + 1 , } \\
{ ( x + 2 ) ^ { 2 } - 3 y ^ { 2 } \geqslant 0 , } \\
{ 2 x + 1 \geqslant 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x^{2}+y^{2} \geqslant 1 \\
(x+2-y \sqrt{3})(x+2+y \sqrt{3}) \geqslant 0 \\
x \geqslant-0.5
\end{array}\right.\right.
$$

The first of these inequalities, together with the second inequality of the original system, defines the set of points located between two concentric circles centered at ( $0 ; 0$ ) with radii 1 and 2. The third inequality defines the half-plane to the right of the line $x=-0.5$. The second inequality defines two vertical angles, the boundaries of which are the lines $\ell_{1}$ and $\ell_{2}$ with equations $y= \pm \frac{x+2}{\sqrt{3}}$ (such that the point $(0 ; 0)$ lies inside one of these angles). The lines $\ell_{1}$ and $\ell_{2}$ both pass through the point ( $-2 ; 0$ ), which lies on the larger circle, and are tangent to the smaller circle at the points $\left(-\frac{1}{2} ; \pm \frac{\sqrt{3}}{2}\right)$ (the coordinates of the points of tangency can be found by solving the corresponding systems of equations).

By intersecting all the specified sets, we obtain the figure $\Phi$, which, as can be easily seen, consists of one part.

## TICKET 14