Given two circles intersecting at points $A$ and $B$. Please construct a line segment $P Q$ passing through point $A$ and intersecting the two circles at points $P$ and $Q$, such that $P Q$ is the longest.

Construction: Let $\odot O_{1}$ and $\odot O_{2}$ intersect at points $A$ and $B$. Connect $B O_{1}$ and let it intersect $\odot O_{1}$ at point $P_{0}$. Connect $P_{0} A$ and let it intersect $\odot O_{2}$ at point $Q_{0}$. Then $P_{0} Q_{0}$ is the desired line segment.
Proof As shown in Figure 1, let the radii of $\odot \mathrm{O}_{1}$ and $\odot \mathrm{O}_{2}$ be $r_{1}$ and $r_{2}$, respectively, and $\left|O_{1} O_{2}\right|=a$. Establish a complex plane with $O_{1} O_{2}$ as the real axis.
For any line segment $P Q$ passing through point $A$, we have
$$
\angle B O_{1} P=2 \angle B A P=\angle B O_{2} Q=t \text {. }
$$

Let the complex numbers representing points $P$, $Q$, and $B$ be $p$, $q$, and $b$, respectively.
Let $b=r_{1} \mathrm{e}^{\mathrm{i} \theta_{1}}=a+r_{2} \mathrm{e}^{\mathrm{i} \theta_{2}}$, i.e.,
$$
r_{2} \mathrm{e}^{\mathrm{i} \theta_{2}}-r_{1} \mathrm{e}^{\mathrm{i} \theta_{1}}=-a \text {. }
$$

Then $p=r_{1} \mathrm{e}^{\mathrm{i}\left(\theta_{1}-t\right)}, q=a+r_{2} \mathrm{e}^{\mathrm{i}\left(\theta_{2}-t\right)}$.
$$
\begin{array}{l}
\text { Hence }|\overrightarrow{P Q}|=\left|a+r_{2} \mathrm{e}^{\mathrm{i}\left(\theta_{2}-t\right)}-r_{1} \mathrm{e}^{\mathrm{i}\left(\theta_{1}-t\right)}\right| \\
=\left|a \mathrm{e}^{\mathrm{i} t}+r_{2} \mathrm{e}^{\mathrm{i} \theta_{2}}-r_{1} \mathrm{e}^{\mathrm{i} \theta_{1}}\right| \\
=\left|a \mathrm{e}^{\mathrm{i} t}-a\right|=a\left|\mathrm{e}^{\mathrm{i} t}-1\right| \\
=a \sqrt{(\cos t-1)^{2}+\sin ^{2} t} \\
=a \sqrt{2-2 \cos t}=2 a \sin \frac{t}{2} .
\end{array}
$$

Therefore, when $t=\pi$, i.e., $\angle B O_{1} P=\pi$,
$$
|\overrightarrow{P Q}|_{\max }=2 a \text {. }
$$
(Ting Ye, Tangxia High School, Ruian City, Zhejiang Province, 325004)