7. Given $F_{1}, F_{2}$ are the left and right foci of the hyperbola $C: \frac{x^{2}}{4}-\frac{y^{2}}{12}=1$, point $P$ is on the hyperbola $C$, and $G, I$ are the centroid and incenter of $\triangle F_{1} P F_{2}$, respectively. If $G I \| x$-axis, then the circumradius $R$ of $\triangle F_{1} P F_{2}$ is $\qquad$ ـ.
Answer 5.
Analysis Suppose $P\left(x_{0}, y_{0}\right)$ is in the first quadrant, $\left|P F_{1}\right|=r_{1},\left|P F_{2}\right|=r_{2}$. According to the problem, $r_{1}-r_{2}=4,\left|F_{1} F_{2}\right|=8$. Since $G, I$ are the centroid and incenter of $\triangle F_{1} P F_{2}$ respectively, and $G I \| x$-axis, the inradius of $\triangle F_{1} P F_{2}$ is $r=\frac{1}{3} y_{0}$, so
$$
S_{\triangle F_{1} P F_{2}}=\frac{1}{2}\left(\left|F_{1} P\right|+\left|F_{1} F_{2}\right|+\left|F_{2} P\right|\right) \cdot r=\frac{1}{2}\left(r_{1}+r_{2}+8\right) \cdot \frac{1}{3} y_{0}
$$

Also, $S_{\triangle F_{1} P F_{2}}=\frac{1}{2} \cdot\left|F_{1} F_{2}\right| \cdot y_{0}=4 y_{0}$, so $\frac{1}{2}\left(r_{1}+r_{2}+8\right) \cdot \frac{1}{3} y_{0}=4 y_{0}$, thus $r_{1}+r_{2}=16$. Combining $r_{1}-r_{2}=4$, we get $r_{1}=10, r_{2}=6$, from which we can conclude that $\left|F_{1} P\right|^{2}=\left|F_{1} F_{2}\right|^{2}+\left|F_{2} P\right|^{2}$, hence $P F_{2} \perp F_{1} F_{2}$. Therefore, the circumradius of $\triangle F_{1} P F_{2}$ is $R=\frac{1}{2}\left|F_{1} P\right|=5$.