9. Specify all values of $a$ for which the system of equations $(x-a)^{2}=16(y-x+a-3), \log _{(x / 3)}(y / 3)=1$ has at least one solution, and solve it for each $a$.
Solution. The second equation is equivalent to the system: $x>0, x \neq 3, y=x$. Substituting $y=x$ into the first equation, we get; $(x-a)^{2}=16(a-3)$, or $x^{2}-2 a x+a^{2}-16 a+48=0(*)$, which has $D / 4=a^{2}-a^{2}+16 a-48=16(a-3)$. The number of solutions to the given system of equations depends on the number of roots of this quadratic equation.

The root $x=3$ of the quadratic equation can occur when $(3-a)^{2}=16(a-3)$, i.e., if 1) $a=3$, the equation becomes $(x-3)^{2}=0$, then this root is the only one and the given system has no solutions, or 2) $3-a=-16$, i.e., $a=19$, then for $x$ we get the equation $(x-19)^{2}=16 \cdot 16, x=19 \pm 16$, which, in addition to the extraneous root $x_{1}=3$, has another root $x_{2}=35$, satisfying the conditions, and the given system has a unique solution $(35 ; 35)$.

Let's consider the other cases when the system will have a unique solution.