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

$$
\left\{\begin{array}{l}
a^{2}-2 a x+10 y+x^{2}+y^{2}=0 \\
(|x|-12)^{2}+(|y|-5)^{2}=169
\end{array}\right.
$$

has exactly two solutions.

![](https://cdn.mathpix.com/cropped/2024_05_06_734640e0e89abd1c02f1g-16.jpg?height=1063&width=1579&top_left_y=882&top_left_x=250)

Fig. 4: variant 10, problem 7
Answer: $a \in(-30 ;-20) \cup\{0\} \cup(20 ; 30)$.

Solution. Consider the second equation of the system. For $x \geqslant 0$ and $y \geqslant 0$ it takes the form $(x-12)^{2}+(y-5)^{2}=169$, and we obtain a circle of radius 13 centered at the point $(12 ; 5)$ (more precisely, its part lying in the first quadrant). Since the equation is invariant under the substitution of $x$ with $(-x)$ and/or $y$ with $(-y)$, the set of points defined by the equation of the system is symmetric with respect to both coordinate axes. Reflecting the obtained arc relative to both coordinate axes and the point $(0 ; 0)$, we get the desired set. Note that it consists of four arcs of circles and the origin.

The first equation can be written in the form $(x-a)^{2}+(y+5)^{2}=25$. It defines a circle of radius 5 centered at the point $(a ;-5)$. Depending on $a$, the center of this circle moves along the line $y=-5$. The system has two solutions if and only if this circle has exactly two common points with the set defined by the second equation. This is possible (see the drawing) when $a \in(-30 ;-20) \cup\{0\} \cup(20 ; 30)$.

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 reduced to 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.

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