9. Specify all values of $a$ for which the system of equations $(x-a)^{2}=8(2 y-x+a-2), \frac{1-\sqrt{y}}{1-\sqrt{x / 2}}=1$ has at least one solution, and solve it for each $a$.

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# Solution:

The second equation is equivalent to the system: $x \geq 0, x \neq 2, y=x / 2$. Substituting $y=x / 2$ into the first equation, we get: $(x-a)^{2}=8(a-2)$, or $x^{2}-2 a x+a^{2}-8 a+16=0(*)$, which has $D / 4=a^{2}-a^{2}+8 a-16=8 a-16$. The number of solutions to the given system of equations depends on the number of roots of this quadratic equation.

Obviously, when $a 0 , } \\
{ a > 0 , } \\
{ a ^ { 2 } - 8 a + 1 6 \geq 0 , } \\
{ 4 - 4 a + a ^ { 2 } - 8 a + 1 6 \neq 0 }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ a > 2 , } \\
{ a ^ { 2 } - 1 2 a + 2 0 \neq 0 }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ a > 2 , } \\
{ a \neq 2 , } \\
{ a \neq 1 0 }
\end{array} \Leftrightarrow \left[\begin{array}{c}
2<a<10 \\
10<a<+\infty
\end{array}\right.\right.\right.\right.
$$

Consider the equation at the remaining individual points.

When $a=2$, the equation becomes $(x-2)^{2}=0$, its root $x=2$ does not satisfy the restrictions imposed on the unknown $x$ in the second equation of the system.

When $a=10$, we get $(x-10)^{2}=8 \cdot 8$, from which $x_{1,2}=10 \pm 8$; the root $x=2$ is extraneous, the second root $x=18$ falls within the domain of admissible values of the system. When $a=10$, the system of equations has one solution $x=18, y=9$.

Combining the found values of $a$, we get the answer.

Answer: $\quad a \in(2 ; 10) \cup(10 ;+\infty), x=a \pm \sqrt{8 a-16}, y=(a \pm \sqrt{8 a-16}) / 2$;

$$
a=10, x=18, y=9
$$