6. Given that $P$ is a point on the hyperbola $C: \frac{x^{2}}{4}-\frac{y^{2}}{12}=1$, $F_{1}, F_{2}$ are the left and right foci of the hyperbola $C$, $M, I$ are the centroid and incenter of $\triangle P F_{1} F_{2}$, respectively. If $M I \perp x$ axis, then the radius of the incircle of $\triangle P F_{1} F_{2}$ is
As shown in the figure, it is easy to know that the incenter $I$ of $\triangle P F_{1} F_{2}$ lies on the line $x= \pm 2$. Without loss of generality, assume point $P$ is in the first quadrant, then $x_{G}=2$. From
$\frac{P G}{G O}=2 \Rightarrow x_{P}=6, y_{P}=4 \sqrt{6}$, thus $\left|P F_{1}\right|=14$,
$\left|P F_{2}\right|=10,\left|F_{1} F_{2}\right|=8, S_{\triangle P F_{1} F_{2}}=16 \sqrt{6}$.
Therefore, $\frac{1}{2}(14+10+8) \cdot r=16 \sqrt{6} \Rightarrow r=\sqrt{6}$.