2. In $\triangle A B C$, the three sides are $a, b, c$, and point $O$ is the circumcenter of $\triangle A B C$. Given that $b^{2}-2 b+c^{2}=0$, find the range of $B C \cdot A O$.

The three sides of $\triangle A B C$ are $a, b, c$, and point $O$ is the circumcenter of $\triangle A B C$. Given that $b^{2}-2 b+c^{2}=0$, the range of $B C \cdot A O$ is
Let $D$ be the midpoint of $AB$, then $\overrightarrow{AB} \cdot \overrightarrow{AO} = \overrightarrow{AB} \cdot (\overrightarrow{AD} + \overrightarrow{DO}) = \frac{1}{2} c^{2}$.
Thus, $\overrightarrow{BC} \cdot \overrightarrow{AO} = \overrightarrow{AO} \cdot (\overrightarrow{AC} - \overrightarrow{AB}) = \frac{1}{2} (b^{2} - c^{2}) = \frac{1}{2} (b^{2} - 2b + b^{2})$
$= b^{2} - b = \left(b - \frac{1}{2}\right)^{2} - \frac{1}{4}$. Since $c^{2} = 2b - b^{2} > 0 \Rightarrow 0 < b < 2 \Rightarrow \overrightarrow{BC} \cdot \overrightarrow{AO} \in \left[-\frac{1}{4}, 2\right)$.