4. As shown in Figure 1, $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $\frac{x^{2}}{16}-\frac{y^{2}}{9}=1$, respectively, and $P$ is a point on the right branch of the hyperbola (excluding the right vertex). $\odot A$ is tangent to the side $P F_{2}$ of $\triangle P F_{1} F_{2}$, and also tangent to the extensions of sides $F_{1} F_{2}$ and $F_{1} P$. As point $P$ moves along the right branch of the hyperbola, the equation of the locus of the center $A$ is $\qquad$
4. $\frac{x^{2}}{25}-\frac{9 y^{2}}{25}=1(x>5)$.

Connecting $P A$ and extending it to intersect the $x$-axis at point $B$, by the properties of the angle bisector, the ratio theorem, and the definition of a hyperbola, we get
$$
\begin{array}{l}
\frac{|P A|}{|A B|}=\frac{\left|P F_{1}\right|}{\left|F_{1} B\right|}=\frac{\left|P F_{2}\right|}{\left|F_{2} B\right|} \\
=\frac{\left|P F_{1}\right|-\left|P F_{2}\right|}{\left|F_{1} B\right|-\left|F_{2} B\right|}=\frac{2 a}{2 c}=\frac{4}{5} .
\end{array}
$$

Let $A(x, y)$ and $P\left(x_{0}, y_{0}\right)\left(x_{0}>4\right)$.
By the section formula, we get
$$
x_{0}=\frac{9 x-4 x_{B}}{5}, y_{0}=\frac{9}{5} y \text {. }
$$

By the focal radius formula, we get
$$
\left|P F_{2}\right|=\frac{5}{4} x_{0}-4 \text {. }
$$

Since $\left|F_{2} B\right|=x_{B}-5$, we have
$$
\frac{\frac{5}{4} x_{0}-4}{x_{B}-5}=\frac{4}{5} \text {. }
$$

Solving for $x_{B}$, we get $x_{B}=\frac{25}{16} x_{0}$.
Thus, $P\left(\frac{4}{5} x, \frac{9}{5} y\right)$.
Substituting into $\frac{x_{0}^{2}}{16}-\frac{y_{0}^{2}}{9}=1$, we get the equation of the locus of the center $A$ as
$$
\frac{x^{2}}{25}-\frac{9 y^{2}}{25}=1(x>5) \text {. }
$$