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 $a$.


9. Let the function $f(x)=x^{2}-2 a x-2 a \ln x$, then $f^{\prime}(x)=2 x-2 a-\frac{2 a}{x}=\frac{2}{x}\left(x^{2}-a x-a\right)$.

Let $f^{\prime}(x)=0$, i.e., $x^{2}-a x-a=0$. From this, we can solve for $x_{1}=\frac{a-\sqrt{a^{2}+4 a}}{2}0$, $f(x)$ is monotonically increasing. Therefore, $f(x)$ reaches its minimum value $f\left(x_{2}\right)$ at $x=x_{2}$.
From the fact that $f(x)=0$ has a unique solution, we know that $f\left(x_{2}\right)=0$, i.e., at this point,
$$
\left\{\begin{array}{l}
f\left(x_{2}\right)=0 \\
f^{\prime}\left(x_{2}\right)=0
\end{array}\right.
$$

That is,
$$
\left\{\begin{array}{l}
x_{2}^{2}-2 a x_{2}-2 a \ln x_{2}=0, \\
x_{2}^{2}-a x_{2}-a=0 .
\end{array}\right.
$$

Thus, $2 a \ln x_{2}-a x_{2}-a=0$. Given $a>0$, we have $2 a \ln x_{2}+x_{2}-1=0$.
Consider the function $g(x)=2 \ln x+x-1$. When $x>0$, $g^{\prime}(x)>0$, so $g(x)$ is monotonically increasing, meaning the equation $g(x)=0$ has at most one solution in $(0,+\infty)$. Noting that $g(1)=0$, we have $x_{2}=1$, from which we can solve for $a=\frac{1}{2}$.