7. Let $F_{1}$ and $F_{2}$ be the left and right foci of the hyperbola $C: \frac{x^{2}}{4}-\frac{y^{2}}{12}=1$, respectively. Point $P$ lies on the hyperbola $C$, and $G$ and $I$ 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 $\qquad$
7.5.

Assume $P\left(x_{0}, y_{0}\right)$ is in the first quadrant. Then $\left|P F_{1}\right|=r_{1},\left|P F_{2}\right|=r_{2}$.
According to the problem, $r_{1}-r_{2}=4,\left|F_{1} F_{2}\right|=8$.
From the given information, the inradius of $\triangle F_{1} P F_{2}$ is $r=\frac{1}{3} y_{0}$.
Thus, $S_{\triangle F_{1} P F_{2}}=\frac{1}{2}\left(\left|F_{1} P\right|+\left|F_{1} F_{2}\right|+\left|F_{2} P\right|\right) r$
$$
=\frac{1}{2}\left(r_{1}+r_{2}+8\right) \frac{1}{3} y_{0} \text {. }
$$

Also, $S_{\triangle F_{1} P F_{2}}=\frac{1}{2}\left|F_{1} F_{2}\right| y_{0}=4 y_{0}$, so,
$$
\begin{array}{l}
\frac{1}{2}\left(r_{1}+r_{2}+8\right) \frac{1}{3} y_{0}=4 y_{0} . \\
\Rightarrow r_{1}+r_{2}=16 .
\end{array}
$$

Combining $r_{1}-r_{2}=4$, we get $r_{1}=10, r_{2}=6$.
Thus, $\left|F_{1} P\right|^{2}=\left|F_{1} F_{2}\right|^{2}+\left|F_{2} P\right|^{2}$.
Hence, $P F_{2} \perp F_{1} F_{2}$.
Therefore, $R=\frac{1}{2}\left|F_{1} P\right|=5$.