7. The equation of the hyperbola $C$ is $x^{2}-\frac{y^{2}}{3}=1$, with the left and right foci being $F_{1}$ and $F_{2}$, respectively. A line is drawn through point $F_{2}$ intersecting the right branch of the hyperbola $C$ at points $P$ and $Q$, such that $\angle F_{1} P Q=90^{\circ}$. Then, the radius of the incircle of $\triangle F_{1} P Q$ is
Answer: $\sqrt{7}-1$.
Solution: By the properties of a hyperbola, we know that $F_{1} F_{2}=2 \times \sqrt{1+3}=4$, $P F_{1}-P F_{2}=Q F_{1}-Q F_{2}=2$.
Since $\angle F_{1} P Q=90^{\circ}$, it follows that $P F_{1}^{2}+P F_{2}^{2}=F_{1} F_{2}^{2}$, thus
$$
\begin{aligned}
P F_{1}+P F_{2} & =\sqrt{2\left(P F_{1}^{2}+P F_{2}^{2}\right)-\left(P F_{1}-P F_{2}\right)^{2}} \\
& =\sqrt{2 \times 4^{2}-2^{2}}=2 \sqrt{7} .
\end{aligned}
$$

Therefore, the inradius of the right triangle $\triangle F_{1} P Q$ is
$$
r=\frac{1}{2}\left(F_{1} P+P Q-F_{1} Q\right)=\frac{1}{2}\left(P F_{1}+P F_{2}\right)-\frac{1}{2}\left(Q F_{1}-Q F_{2}\right)=\sqrt{7}-1 .
$$