Question 167, Given the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$ with left and right foci $F_{1}$ and $F_{2}$, respectively, a line $l$ is drawn through the right focus of the ellipse, intersecting the ellipse at points $\mathrm{P}$ and $\mathrm{Q}$. Try to find the maximum value of the area of the incircle of $\triangle \mathrm{F}_{1} \mathrm{PQ}$.
Question 167, Answer: According to the conditions, the semi-major axis of the ellipse is $a=2$, the semi-minor axis is $b=\sqrt{3}$, the semi-focal distance is $c=1$, the eccentricity $\mathrm{e}=\frac{1}{2}$, and $\mathrm{p}=\frac{\mathrm{a}^{2}}{\mathrm{c}}-\mathrm{c}=3$. The perimeter of $\triangle \mathrm{F}_{1} \mathrm{PQ}$ is $4 a=8$. Establishing a polar coordinate system with the right focus as the pole, the polar equation is $\rho=\frac{\mathrm{ep}}{1+\mathrm{e} \cdot \cos \theta}=\frac{\frac{3}{2}}{1+\frac{\cos \theta}{2}}=\frac{3}{2+\cos \theta}$. Therefore, the area of $\triangle F_{1} P Q$ is
$$
S_{\triangle F_{1} P Q}=\left|\frac{\left(\frac{3}{2+\cos \theta} \frac{3}{2-\cos \theta}\right) \cdot 2 \sin \theta}{2}\right|=\left|\frac{12 \sin \theta}{4-\cos ^{2} \theta}\right|=\frac{12}{\left|\frac{3}{\sin \theta}+\sin \theta\right|} \leq \frac{12}{4}=3
$$

Thus, the inradius of $\triangle F_{1} P Q$ is $r=\frac{S_{\Delta F_{1} P Q}}{\frac{4 a}{2}} \leq \frac{3}{4}$, so the maximum value of its inradius is $\pi r^{2}=\frac{9 \pi}{16} \circ$