4. Connect the right focus $F$ of the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$ with a moving point $A$ on the ellipse, and construct a square $F A B C (F, A, B, C$ in clockwise order). When point $A$ moves around the ellipse once, the trajectory equation of the moving point $C$ is $\qquad$
4. $\frac{(x-\sqrt{5})^{2}}{4}+\frac{(y-\sqrt{5})^{2}}{9}=1$.

Let $A(3 \cos \alpha, 2 \sin \alpha)(\alpha \in[0,2 \pi])$,
$$
\begin{aligned}
C(x, y) 、 F(\sqrt{5}, 0), \\
\overrightarrow{F A}=(3 \cos \alpha-\sqrt{5}, 2 \sin \alpha) . \\
\text { Then } \overrightarrow{F C}=(2 \sin \alpha,-3 \cos \alpha+\sqrt{5}) . \\
\text { Also } \overrightarrow{F C}=(x-\sqrt{5}, y), \\
\left\{\begin{array}{l}
x=\sqrt{5}+2 \sin \alpha, \\
y=\sqrt{5}-3 \cos \alpha .
\end{array}\right.
\end{aligned}
$$

Eliminating $\alpha$ yields the trajectory equation of the moving point $C$ as
$$
\frac{(x-\sqrt{5})^{2}}{4}+\frac{(y-\sqrt{5})^{2}}{9}=1 .
$$