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

$$
\left\{\begin{array}{l}
a^{2}-2 a x-6 y+x^{2}+y^{2}=0 \\
(|x|-4)^{2}+(|y|-3)^{2}=25
\end{array}\right.
$$

has exactly two solutions.
Answer: $a \in(-12 ;-6) \cup\{0\} \cup(6 ; 12)$.

Solution. Consider the second equation of the system. For $x \geqslant 0$ and $y \geqslant 0$, it takes the form $(x-4)^{2}+(y-3)^{2}=25$, and we obtain a circle of radius 5 centered at the point $(4 ; 3)$ (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 as $(x-a)^{2}+(y-3)^{2}=9$. It defines a circle of radius 3 centered at the point $(a ; 3)$. Depending on $a$, the center of this circle moves along the line $y=3$. 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 $a \in(-12 ;-6) \cup\{0\} \cup(6 ; 12)$.

![](https://cdn.mathpix.com/cropped/2024_05_06_734640e0e89abd1c02f1g-13.jpg?height=1050&width=1585&top_left_y=246&top_left_x=247)

Fig. 2: variant 9, problem 7

## VARIANT 10