2. Given, $\odot \mathrm{O}_{1}$ and $\odot \mathrm{O}_{2}$ intersect at $\mathrm{A}$ and $\mathrm{B}$, a tangent line $\mathrm{AC}$ is drawn from point $\mathrm{A}$ to $\odot \mathrm{O}_{2}$, $\angle \mathrm{CAB}=45^{\circ}$, the radius of $\odot \mathrm{O}_{2}$ is $5 \sqrt{2} \mathrm{~cm}$, find the length of $\mathrm{AB}$ (Figure 2).
2. Solution: As shown in the figure

Draw diameter $\mathrm{AD}$, and connect $\mathrm{BD}$.
$$
\begin{array}{c}
\because \mathrm{AC} \text { is tangent to } \odot \text { at } \mathrm{A}, \text { and } \because \angle \mathrm{CAB}=45^{\circ}, \\
\therefore \angle \mathrm{D}=\angle \mathrm{CAB}=45^{\circ} . \\
\mathrm{AO}_{2}=5 \sqrt{2} \mathrm{~cm}, \therefore \mathrm{AD}=10 \sqrt{2} \mathrm{~cm} . \\
\mathrm{AD} \text { is the diameter of } \odot \mathrm{O}, \therefore \angle \mathrm{ABD}=90^{\circ} . \\
\therefore \mathrm{AB}=\mathrm{BD}=\sqrt{\frac{(10 \sqrt{2})^{2}}{2}}=10(\mathrm{~cm}) .
\end{array}
$$