6. Find all values of the parameter $a$ for which the system $\left\{\begin{array}{l}5|x|+12|y|=60, \\ x^{2}+y^{2}-2 y+1-a^{2}=0\end{array} \quad\right.$ a) has exactly 3 solutions; b) has exactly 2 solutions.
Answer. a) $|a|=4$; b) $|a| \in\left\{\frac{48}{13}\right\} \cup\left(4 ; \frac{72}{13}\right) \cup\{\sqrt{145}\}$.

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 $y=4-\frac{4}{3} x$ - a segment connecting the points $(12 ; 0)$ and $(0 ; 5)$. Using the symmetry of the set with respect to the coordinate axes, we obtain a rhombus with vertices $A(0 ; 5), B(-12 ; 0)$, $C(0 ;-5), D(12 ; 0)$.

The second equation of the system can be written as $x^{2}+(y-1)^{2}=a^{2}$. It defines a circle with center $Q(0 ; 1)$ 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 rhombus and the circle are symmetric with respect to the y-axis, so 3 solutions are possible only if one of the common points of the circle and the rhombus 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., $|a|=4$ or $|a|=6$. It is easy to see that when $|a|=4$, the system has 3 solutions, and when $|a|=6$, there are 5 solutions. Therefore, 3 solutions are possible only when $a= \pm 4$.

b) Let $R_{0}$ be the radius of the circle that is tangent to sides $B C$ and $C D$, and $R_{1}$ be the radius of the circle that is tangent to sides $A B$ and $A D$ of the rhombus. The system has exactly two solutions if and only if $|a| \in\left\{R_{1}\right\} \cup\left(Q A ; R_{0}\right) \cup\{Q B\}$.

$Q A=4, Q B=\sqrt{12^{2}+1^{2}}=\sqrt{145}$. Let the circle of radius $R_{1}$ touch side $A B$ at point $J$, and the circle of radius $R_{0}$ touch side $B C$ at point $L$. Triangle $J A Q$ is a right triangle, $\angle J Q A=90^{\circ}-\angle J A Q=\angle A B D$, so $\operatorname{tg} \angle J Q A=\operatorname{tg} \angle A B D=\frac{5}{12}$, since it is equal to the slope of the line $A B$. Then $A J=J Q \operatorname{tg} \angle J Q A=\frac{5 R_{1}}{12}$. By the Pythagorean theorem for triangle $J Q A$, we get $16=R_{1}^{2}+\frac{25 R_{1}^{2}}{144}$, from which $R_{1}=\frac{48}{13}$. Since triangles $J Q A$ and $L Q C$ are similar and the similarity ratio is $\frac{Q A}{Q C}=\frac{2}{3}$, then $R_{0}=Q L=\frac{3}{2} Q J=\frac{72}{13}$. Finally, we get $|a| \in\left\{\frac{48}{13}\right\} \cup\left(4 ; \frac{72}{13}\right) \cup\{\sqrt{145}\}$.