Example 1 Let $P$ and $Q$ be two moving points on the circumference of the circle $x^{2}+y^{2}=1$, and they satisfy the condition that with a fixed point $A\left(0, \frac{1}{2}\right)$ inside the circle, $\angle P A Q = \frac{\pi}{2}$. Find the equation of the locus of point $M$, the intersection of the two tangents passing through points $P$ and $Q$.
(2007, Zhejiang Province High School Mathematics Competition (B Volume))
Solution 1: Let $M\left(x_{0}, y_{0}\right)$. Then the equation of the line containing the chord $P Q$ is $x_{0} x+y_{0} y=1$.
Substituting into $x^{2}+y^{2}=1$ gives
$$
\left(x_{0}^{2}+y_{0}^{2}\right) x^{2}-2 x_{0} x+1-y_{0}^{2}=0 \text {. }
$$

Let $P\left(x_{1}, y_{1}\right)$ and $Q\left(x_{2}, y_{2}\right)$. Then
$$
\begin{array}{l}
x_{1}+x_{2}=\frac{2 x_{0}}{x_{0}^{2}+y_{0}^{2}}, x_{1} x_{2}=\frac{1-y_{0}^{2}}{x_{0}^{2}+y_{0}^{2}}, \\
y_{1}+y_{2}=\frac{2 y_{0}}{x_{0}^{2}+y_{0}^{2}}, y_{1} y_{2}=\frac{1-x_{0}^{2}}{x_{0}^{2}+y_{0}^{2}} .
\end{array}
$$

Since $\angle P A Q=\frac{\pi}{2}$, then
$$
x_{1} x_{2}+\left(y_{1}-\frac{1}{2}\right)\left(y_{2}-\frac{1}{2}\right)=0 \text {. }
$$

Substituting and simplifying gives $3 x_{0}^{2}+3 y_{0}^{2}+4 y_{0}-8=0$.
Thus, the equation of the locus of point $M$ is
$$
3 x^{2}+3 y^{2}+4 y-8=0 \text {. }
$$

Solution 2: As shown in Figure 1,
Notice that
$$
\begin{array}{l}
O P=O Q, \\
M P=M Q, \\
\angle P A Q=\frac{\pi}{2} .
\end{array}
$$

Connecting $O M$ to intersect $P Q$ at $R$, then $O M$ perpendicularly bisects $P Q$.
In the right triangle $\triangle P A Q$, $A R=\frac{1}{2} P Q=P R$.
In the right triangle $\triangle O P R$, $O R^{2}+P R^{2}=O P^{2}$, i.e.,
$$
O R^{2}+A R^{2}=1 \text {. }
$$

Let $M\left(x_{0}, y_{0}\right)$. Then the equation of the line containing the chord $P Q$ is $x_{0} x+y_{0} y=1$.
Substituting into $x^{2}+y^{2}=1$ gives
$$
\left(x_{0}^{2}+y_{0}^{2}\right) x^{2}-2 x_{0} x+1-y_{0}^{2}=0 \text {. }
$$

Let $P\left(x_{1}, y_{1}\right)$, $Q\left(x_{2}, y_{2}\right)$, and $R\left(x_{3}, y_{3}\right)$. Then
$$
\begin{array}{l}
x_{3}=\frac{x_{1}+x_{2}}{2}=\frac{x_{0}}{x_{0}^{2}+y_{0}^{2}}, \\
y_{3}=\frac{y_{1}+y_{2}}{2}=\frac{y_{0}}{x_{0}^{2}+y_{0}^{2}},
\end{array}
$$

Substituting into equation (1) and simplifying gives
$$
\frac{2}{x_{0}^{2}+y_{0}^{2}}-\frac{y_{0}}{x_{0}^{2}+y_{0}^{2}}+\frac{1}{4}=1,
$$

i.e., $3 x_{0}^{2}+3 y_{0}^{2}+4 y_{0}-8=0$.
Thus, the equation of the locus of point $M$ is
$$
3 x^{2}+3 y^{2}+4 y-8=0 \text {. }
$$