7. Find all values of the parameter $a$ for which the system

$$
\left\{\begin{array}{l}
3|y|-4|x|=6 \\
x^{2}+y^{2}-14 y+49-a^{2}=0
\end{array}\right.
$$

a) has exactly 3 solutions; b) has exactly 2 solutions.
Answer. a) $|a| \in\{5 ; 9\} ;$ b) $|a| \in\{3\} \cup(5 ; 9)$.

Solution. The first equation of the system does not change when $x$ is replaced by $-x$ and/or $y$ by $-y$. Therefore, the set of points defined by the first equation is symmetric with respect to both coordinate axes. In the first quadrant, we get a part of the line $y=2+\frac{4}{3} x$ - a ray starting at the point $(0 ; 2)$ with a slope of $\frac{4}{3}$. Using the symmetry of the set with respect to the coordinate axes, we obtain 2 angles: one with the vertex at point $A(0 ; 2)$ with branches pointing upwards, and the other - with the vertex at $C(0 ;-2)$ and branches pointing downwards (the slopes of the sides of the angle are $\pm \frac{4}{3}$ ).

The second equation of the system can be written as $x^{2}+(y-7)^{2}=a^{2}$. It defines a circle with center $Q(0 ; 7)$ and radius $|a|$ (or the point $Q$ if $a=0$ ). If $a=0$, there are no solutions, so let's consider the case of a circle.

a) Both the circle and the set of points defined by the first equation are symmetric with respect to the y-axis, so 3 solutions are possible only if one of their common points lies on the y-axis. This happens if the radius of the circle is equal to the segment $Q A$ or the segment $Q C$, i.e., $|a|=5$ or $|a|=9$. It is easy to see that for these $a$, the circle has two more common points with the sides of the angle lying in the upper half-plane, and the system has a total of 3 solutions. Then $a= \pm 5$ or $a= \pm 9$.
b) Let $R_{0}$ be the radius of the circle that is tangent to the sides of the angle with vertex $A$. The system has two solutions when $|a| \in\left\{R_{0}\right\} \cup(Q A ; Q C)$.

Drop a perpendicular $Q H$ from point $Q$ to the side of the angle lying in the first quadrant. Let $\alpha$ be the angle of inclination of the line $A H\left(\operatorname{tg} \alpha=\frac{4}{3}\right)$. Then $\angle Q A H=90^{\circ}-\alpha, \angle A Q H=\alpha$. Since $Q H=R_{0}$, then $A H=Q H \operatorname{tg} \alpha=\frac{4}{3} R_{0}$ and by the Pythagorean theorem for triangle $A Q H$ we get $R_{0}^{2}+\frac{16}{9} R_{0}^{2}=25$, from which $R_{0}=3$. Therefore, $|a| \in\{3\} \cup(5 ; 9)$.

## TICKET 16