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

$$
\left\{\begin{array}{l}
\sqrt{y^{2}-8 x^{2}-6 y+9} \leqslant 3 y-1 \\
x^{2}+y^{2} \leqslant 9
\end{array}\right.
$$

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

$$
\left\{\begin{array} { l } 
{ y ^ { 2 } - 8 x ^ { 2 } - 6 y + 9 \leqslant 9 y ^ { 2 } - 6 y + 1 , } \\
{ ( y - 3 ) ^ { 2 } - 8 x ^ { 2 } \geqslant 0 , } \\
{ 3 y - 1 \geqslant 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x^{2}+y^{2} \geqslant 1 \\
(y-3-2 x \sqrt{2})(y-3+2 x \sqrt{2}) \geqslant 0 \\
y \geqslant \frac{1}{3}
\end{array}\right.\right.
$$

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