5. Find all values of the parameter $a$, for each of which there exists a number $b$ such that the system

$$
\left\{\begin{array}{l}
x=|y+a|+\frac{4}{a} \\
x^{2}+y^{2}+24+b(2 y+b)=10 x
\end{array}\right.
$$

has at least one solution $(x ; y)$.
Answer. $a \in(-\infty, 0) \cup\left[\frac{2}{3} ;+\infty\right)$.

Solution. The second equation of the system can be transformed into the form $(x-5)^{2}+(y+b)^{2}=1^{2}$, hence it represents a circle of radius 1 with center $(5 ;-b)$. For all possible $b \in \mathbf{R}$, these circles sweep the strip $4 \leq x \leq 6$.

The first equation represents a "corner" with branches directed to the right, with the vertex at the point $\left(\frac{4}{a} ;-a\right)$.

For the condition of the problem to be satisfied, it is necessary and sufficient that the "corner" defined by the first equation has at least one common point with the strip $4 \leq x \leq 6$, and for this, the abscissa of its vertex must satisfy the inequality $x_{\mathrm{B}} \leq 6$, i.e., $\frac{4}{a} \leq 6$ from which $a \in(-\infty, 0) \cup\left[\frac{2}{3} ;+\infty\right)$.