Example 14 (Question from the 10th "Hope Cup" Invitational Competition) Given that real numbers $x, y$ satisfy the equation $(x+2)^{2}+y^{2}=1$, what is the minimum value of $\frac{y-1}{x-2}$?
Let $\frac{y-1}{x-2}=k$, then $y-1=k x-2 k, y=k x-2 k+1$.
Let $\vec{a}=(x+2, k x-2 k+1), \vec{b}=(k,-1)$,
$$
\begin{array}{l} 
1=(x+2)^{2}+y^{2}=(x+2)^{2}+(k x-2 k+1)^{2} \\
=|\vec{a}|^{2} \geqslant \frac{(\vec{a} \cdot \vec{b})^{2}}{|\vec{b}|^{2}}=\frac{(k x+2 k-k x+2 k-1)^{2}}{k^{2}+1^{2}}=\frac{(4 k-1)^{2}}{k^{2}+1}, \\
(4 k-1)^{2} \leqslant k^{2}+1, k(15 k-8) \leqslant 0,
\end{array}
$$

Therefore, $0 \leqslant k \leqslant \frac{8}{15}$, so the minimum value of $\frac{y-1}{x-2}=k$ is 0.