7. [7 points] Find all values of the parameter $a$, for each of which the system

$$
\left\{\begin{array}{l}
(x-a)^{2}+y^{2}=64 \\
(|x|-8)^{2}+(|y|-15)^{2}=289
\end{array}\right.
$$

has exactly two solutions.
Answer: $a \in\{-28\} \cup(-24 ;-8] \cup[8 ; 24) \cup\{28\}$.

Solution. Consider the second equation of the system. It is invariant under the substitution of $x$ with $(-x)$ and/or $y$ with $(-y)$. This means that the set of points defined by this equation is symmetric with respect to both coordinate axes. In the first quadrant (including its boundaries), by expanding the absolute values, we get $(x-8)^{2}+(y-15)^{2}=289$. This equation defines a circle with center $A(8 ; 15)$ and radius 17. In the first quadrant, there is an arc of this circle and the point $O(0 ; 0)$. By reflecting this arc symmetrically with respect to the origin and both coordinate axes, we obtain the set of points defined by the second equation (denote this set as $M$ - see the figure).

The geometric locus of points defined by the first equation is a circle $\Omega$ with radius 8 and center at point $F(a ; 0)$.

![](https://cdn.mathpix.com/cropped/2024_05_06_c4692c563a78e68b8782g-19.jpg?height=1085&width=1102&top_left_y=243&top_left_x=497)

Fig. 4: variant 12, problem 7

Consider positive values of $a$.

If $0<a<8$, the circle $\Omega$ has no common points with the set $M$.

If $a=8$, we get two solutions (points $(0 ; 0)$ and $(16 ; 0)$).

If $8<a<24$, the system has two solutions.

When $a=24$, the circle $\Omega$ has three common points with the set $M$ (depicted with a dashed line in the figure).

If $24<a<28, \Omega$ intersects the figure $M$ twice in the first quadrant and twice in the fourth quadrant - in total, we have four solutions.

Finally, when $a=28$, $\Omega$ and $M$ have exactly two common points (the case of tangency - see the drawing).

The value $a=28$, at which the circles are tangent, can be found as follows. Let $P(a ; 0)$ be the position of the center of the circle at which the tangency occurs, and $H(8 ; 0)$ be the foot of the perpendicular dropped from point $A$ to the x-axis. Then $A H=15$; the segment $A P$ is the distance between the centers of the tangent circles, so it is equal to the sum of the radii, i.e., $A P=17+8=25$. Then, by the Pythagorean theorem in triangle $A H P$, we get $H P=\sqrt{25^{2}-15^{2}}=20$. Therefore, the point $P$ has an x-coordinate of $8+20=28, a=28$.

Thus, among the positive values of $a$, the suitable ones are $a \in[8 ; 24) \cup\{28\}$. By symmetry, for negative values of the parameter, we get $a \in\{-28\} \cup(-24 ;-8]$. It is also clear that the value $a=0$ does not work. Therefore, $a \in\{-28\} \cup(-24 ;-8] \cup[8 ; 24) \cup\{28\}$.

The problem is considered fully solved (and the maximum number of points is awarded) only if all necessary transformations are provided in the solution text and all logical steps are fully explained; the answers obtained are presented in a simplified form.

The presence of a correct answer does not guarantee a positive score for the problem. A correct answer without justification does not add points.

A full and justified solution to the problem is awarded the full number of points (indicated in parentheses after the problem number). Some partial progress is evaluated according to the instructions. In other cases, the score is at the discretion of the checker.

A single arithmetic error that does not significantly affect the solution process results in a deduction of 1 point.