9. Specify all values of $a$ for which the system of equations $(x-a)^{2}=4(y-x+a-1), \frac{\sqrt{y}-1}{\sqrt{x}-1}=1$ has at least one solution, and solve it for each $a$.

#
# Solution:

The second equation is equivalent to the system: $x \geq 0, x \neq 1, y=x$. Substituting $y=x$ into the first equation, we get: $(x-a)^{2}=4(a-1)$, or $x^{2}-2 a x+a^{2}-4 a+4=0(*)$, for which $D / 4=a^{2}-a^{2}+4 a-4=4(a-1)$. 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 } - 4 a + 4 \geq 0 , } \\
{ 1 - 2 a + a ^ { 2 } - 4 a + 4 \neq 0 }
\end{array} \Leftrightarrow \left\{\begin{array} { r } 
{ a > 1 , } \\
{ a ^ { 2 } - 6 a + 5 \neq 0 }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ a > 1 , } \\
{ a \neq 1 , } \\
{ a \neq 5 }
\end{array} \Leftrightarrow \left[\begin{array}{c}
1<a<5 \\
5<a<+\infty
\end{array}\right.\right.\right.\right.
$$

Consider the equation at the remaining individual points.

When $a=1$, the equation becomes $(x-1)^{2}=0$, its root $x=1$ does not satisfy the restrictions imposed on the unknown $x$ in the second equation of the system.

When $a=5$, we get $(x-5)^{2}=4 \cdot 4$, from which $x_{1,2}=5 \pm 4$; the root $x=1$ is extraneous, the second root $x=9$ falls within the domain of admissible values of the system. When $a=5$, the system of equations has one solution $x=9, y=9$.

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

Answer: $\quad a \in(1 ; 5) \cup(5 ;+\infty), x=y=a \pm 2 \sqrt{a-1}$;

$$
a=5, x=9, y=9
$$