6. Given that $P$ is a point on the hyperbola $C: \frac{x^{2}}{4}-\frac{y^{2}}{12}=1$, $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $C$, and $M$ and $I$ 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 $\qquad$
6. $\sqrt{6}$.

As shown in Figure 3, let point $P$ be in the first quadrant, and $D$, $E$, $F$ be the points where the circle $\odot I$ is tangent to the three sides of $\triangle P F_{1} F_{2}$. Then, by the tangent segment theorem and the definition of a hyperbola, we have
$$
\begin{array}{l}
2 a=\left|P F_{1}\right|-\left|P F_{2}\right| \\
=\left|F F_{1}\right|-\left|E F_{2}\right|=\left|F_{1} D\right|-\left|F_{2} D\right| \\
=\left(x_{D}+c\right)-\left(c-x_{D}\right)=2 x_{D} \\
\Rightarrow x_{D}=a=2, x_{M}=x_{I}=x_{D}=2 .
\end{array}
$$

Let $P\left(x_{0}, y_{0}\right)$.
Since $M$ is the centroid of $\triangle P F_{1} F_{2}$, we have
$$
x_{0}=3 x_{M}=6, y_{0}=4 \sqrt{6} \text {. }
$$
$$
\begin{array}{l}
\text { Then }\left|P F_{1}\right|=\sqrt{(6+4)^{2}+(4 \sqrt{6}-0)^{2}}=14, \\
\left|P F_{2}\right|=\sqrt{(6-4)^{2}+(4 \sqrt{6}-0)^{2}}=10 .
\end{array}
$$

Let the inradius of $\triangle P F_{1} F_{2}$ be $r$. Therefore,
$$
\begin{array}{l}
S_{\triangle P F_{1} F_{2}}=\frac{1}{2}\left(\left|P F_{1}\right|+\left|P F_{2}\right|+\left|F_{1} F_{2}\right|\right) r \\
=16 r . \\
\text { Also, } S_{\triangle P F_{1} F_{2}}=\frac{1}{2}\left|F_{1} F_{2}\right| y_{0} \\
=\frac{1}{2} \times 8 \times 4 \sqrt{6}=16 \sqrt{6},
\end{array}
$$

Thus, $16 r=16 \sqrt{6} \Rightarrow r=\sqrt{6}$.