5. Determine all values of the parameter $p$ for which the system

$$
\left\{\begin{array}{l}
(x-p)^{2}=16(y-3+p) \\
y^{2}+\left(\frac{x-3}{|x|-3}\right)^{2}=1
\end{array}\right.
$$

has solutions and solve it for each of the found values of $p$.

points)
Solution. Domain of definition: $|x| \neq 3$. For $x \geq 0 \quad y^{2}+1=1 \Rightarrow y=0$. If $x0, \\ x_{1} x_{2}=p^{2}-16 p+48=(p-12)(p-4)0, \\ x_{1} x_{2}=p^{2}-16 p+48=(p-12)(p-4) \geq 0, \Rightarrow \\ x_{1}+x_{2}>0\end{array}\right.$

Thus, we get the answer:

$x_{1,2}=p \pm 4 \sqrt{p-3}, p \in(3 ; 4] \cup[12 ;+\infty), \mathrm{p} \neq 19$

$p \in(3 ; 4] \cup[12 ; 19) \bigcup(19 ; \infty), x_{1,2}=p \pm 4 \sqrt{p-3}, \quad y=0$,

$p \in(4 ; 12), p=19$,

$x=p+4 \sqrt{p-3}, \quad y=0$.