9. Let $x, y \in \mathbf{R}$ satisfy $x-6 \sqrt{y}-4 \sqrt{x-y}+12=0$, then the range of values for $x$ is
Given $14-2 \sqrt{13} \leq x \leq 14+2 \sqrt{13}$.
Analysis From $x-6 \sqrt{y}-4 \sqrt{x-y}+12=0 \Rightarrow(\sqrt{x-y}-2)^{2}+(\sqrt{y}-3)^{2}=1$, let $\sqrt{x-y}-2=\cos \theta$,
$$
\sqrt{y}-3=\sin \theta \Rightarrow x=(2+\cos \theta)^{2}+(3+\sin \theta)^{2}=14+\sqrt{52} \sin (\theta+\varphi) \quad\left(\sin \varphi=\frac{2}{\sqrt{13}}\right)
$$

Therefore, $14-2 \sqrt{13} \leq x \leq 14+2 \sqrt{13}$.