Example 7 Let the equation of the ellipse be $\frac{x^{2}}{100}+\frac{y^{2}}{36}=1, P\left(\frac{25}{4}, \frac{3 \sqrt{39}}{4}\right), F_{1}, F_{2}$ are the foci, find the equation of the incircle of $\triangle P F_{1} F_{2}$.
Solution: Clearly, point $P$ is on the ellipse. Let the incenter of $\triangle P F_{1} F_{2}$ be $I$, and $PI$ intersects the $x$-axis at $M$. It is easy to know that $F_{1}(-8, 0), F_{2}(8,0)$, and we can find that $\left|P F_{1}\right|=15, \left|P F_{2}\right|=5$. From $\frac{\left|P F_{1}\right|}{\left|P F_{2}\right|}=\frac{\left|F_{1} M\right|}{\left|M F_{2}\right|}=3$, we get $x_{M}=4$.
By property $4(\mathrm{I})$, we have $\left|\frac{M I}{P I}\right|=e=\frac{4}{5}$.
Thus, $x_{I}=\frac{4+\frac{25}{4} \cdot \frac{4}{5}}{1+\frac{4}{5}}=5, y_{I}=\frac{0+\frac{3}{4} \sqrt{39} \cdot \frac{4}{5}}{1+\frac{4}{5}}=\frac{\sqrt{39}}{3}$.
Since the inradius $r=\left|y_{I}\right|$, the equation of the required circle is
$$
(x-5)^{2}+\left(y-\frac{\sqrt{39}}{3}\right)^{2}=\frac{13}{3} \text {. }
$$