1. Given that $\odot O_{1}$ and $\odot O_{2}$ are disjoint, with radii $r_{1}$ and $r_{2}$ respectively ($r_{1}<r_{2}$), $AB$ and $XY$ are two internal common tangents of the two circles, with points $A$ and $X$ on $\odot O_{1}$, and points $B$ and $Y$ on $\odot O_{2}$. The circle with $AB$ as its diameter intersects $\odot O_{1}$ and $\odot O_{2}$ at the second points $P$ and $Q$ respectively. If $\angle A O_{1} P + \angle B O_{2} Q = 180^{\circ}$, find the value of $\frac{P X}{Q Y}$ (expressed in terms of $r_{1}$ and $r_{2}$).
1. As shown in Figure 1.

Let $A B$ intersect $X Y$ at point $S$. Then $S$ is the internal center of similitude of $\odot O_{1}$ and $\odot O_{2}$, and $\frac{A S}{B S}=\frac{r_{1}}{r_{2}}$.
Let $M$ be the midpoint of $A B$.
Since $\angle A P B=\angle A Q B=90^{\circ}$, we have $M A=M P=M B=M Q$, which implies that $M P$ is tangent to $\odot O_{1}$ and $M Q$ is tangent to $\odot O_{2}$.
From the given conditions, we have
$$
\begin{array}{l}
\angle P O_{1} M=\frac{1}{2} \angle A O_{1} P \\
=\frac{1}{2}\left(180^{\circ}-\angle B O_{2} Q\right) \\
=90^{\circ}-\angle B O_{2} M=\angle B M O_{2} .
\end{array}
$$

Thus, $\triangle O_{1} P M \backsim \triangle M B O_{2}$
$$
\Rightarrow \frac{P O_{1}}{P M}=\frac{B M}{B O_{2}}=\sqrt{\frac{P O_{1}}{P M} \cdot \frac{B M}{B O_{2}}}=\sqrt{\frac{r_{1}}{r_{2}}} \text {. }
$$

Since $\angle P A B=\angle P O_{1} M=\angle B M O_{2}=\angle B A Q$
$\Rightarrow$ points $P$ and $Q$ are symmetric with respect to $A B$.
Let $A B$ intersect $P Q$ at point $S^{\prime}$. Then
$$
\begin{array}{l}
\frac{A S^{\prime}}{S^{\prime} B}=\frac{S_{\triangle A S^{\prime} P}}{S_{\triangle P S^{\prime} B}}=\left(\frac{A P}{B P}\right)^{2}=\cot ^{2} \angle P A B \\
=\cot ^{2} \angle P O_{1} M=\frac{r_{1}}{r_{2}}=\frac{A S}{B S} .
\end{array}
$$

This indicates that point $S^{\prime}$ coincides with $S$, i.e., $P, S, Q$ are collinear.
By the power of a point theorem, we have
$$
\begin{array}{l}
S P \cdot S Q=S A \cdot S B=S X \cdot S Y \\
\Rightarrow \triangle S X P \backsim \triangle S Q Y \\
\Rightarrow \frac{P X}{Q Y}=\frac{S P}{S Y}=\frac{S P}{S B}=\frac{A P}{P B}=\sqrt{\frac{r_{1}}{r_{2}}} .
\end{array}
$$