9. (16 points) Given $a \in \mathbf{R}_{+}$, the equation
$$
x^{2}-2 a x-2 a \ln x=0
$$

has a unique solution in $(0,+\infty)$. Find the value of $a$.
Let the function be
$f(x)=x^{2}-2 a x-2 a \ln x$.
Then $f^{\prime}(x)=2 x-2 a-\frac{2 a}{x}$
$$
=\frac{2}{x}\left(x^{2}-a x-a\right) \text {. }
$$

Let $f^{\prime}(x)=0$, i.e., $x^{2}-a x-a=0$.
Solving this, we get $x_{1}=\frac{a-\sqrt{a^{2}+4 a}}{2}<0$, and $f(x)$ is monotonically increasing.
Thus, $f(x)$ attains its minimum value $f\left(x_{2}\right)$ at $x=x_{2}$.
Given that $f(x)=0$ has a unique solution, we know $f\left(x_{2}\right)=0$, i.e.,
$$
\left\{\begin{array}{l}
x_{2}^{2}-2 a x_{2}-2 a \ln x_{2}=0, \\
x_{2}^{2}-a x_{2}-a=0 .
\end{array}\right.
$$

Therefore, $2 a \ln x_{2}+a x_{2}-a=0$.
Since $a>0$, we have
$2 \ln x_{2}+x_{2}-1=0$.
For $x>0$, the function $g(x)=2 \ln x+x-1$ is strictly increasing, and $g(1)=0$; thus, $x_{2}=1$.
From this, we can solve for $a=\frac{1}{2}$.