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

$$
\left\{\begin{array}{l}
5|x|-12|y|=5 \\
x^{2}+y^{2}-28 x+196-a^{2}=0
\end{array}\right.
$$

a) has exactly 3 solutions; b) has exactly 2 solutions.
Answer. a) $|a| \in\{13 ; 15\} ;$ b) $|a| \in\{5\} \cup(13 ; 15)$.

Solution. The first equation of the system does not change when $x$ is replaced by $-x$ and/or $y$ is replaced 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 $y=\frac{5 x}{12}-\frac{5}{12}-$ a ray starting at the point $(1 ; 0)$ with a slope of $\frac{5}{12}$. Using the symmetry of the set with respect to the coordinate axes, we get 2 angles: one with the vertex at point $A(1 ; 0)$ with branches to the right, and the other - with the vertex at $C(-1 ; 0)$ and branches to the left (the slopes of the sides of the angle are $\pm \frac{5}{12}$ ).

The second equation of the system can be written as $(x-14)^{2}+y^{2}=a^{2}$. It defines a circle with center $Q(14 ; 0)$ and radius $|a|$ (or the point $Q$, if $a=0$ ). When $a=0$, 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 x-axis, so 3 solutions are possible only if one of their common points lies on the x-axis. This happens if the radius of the circle is equal to the segment $Q A$ or the segment $Q C$, i.e., $|a|=13$ or $|a|=15$. 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 right half-plane, and the system has a total of 3 solutions. Then $a= \pm 13$ or $a= \pm 15$.

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 $A H\left(\operatorname{tg} \alpha=\frac{5}{12}\right)$. Then $\angle Q A H=\alpha$. Since $Q H=R_{0}$, we have $A H=\frac{Q H}{\operatorname{tg} \alpha}=\frac{12}{5} R_{0}$ and by the Pythagorean theorem for triangle $A Q H$ we get $R_{0}^{2}+\frac{144}{25} R_{0}^{2}=169$, from which $R_{0}=5$. Therefore, $|a| \in\{5\} \cup(13 ; 15)$.