13.3.23 $\star \star$ Let $P$ and $Q$ be two moving points on the circumference of the circle $x^{2}+y^{2}=1$, and 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 locus of the intersection point $M$ of the two tangents passing through $P$ and $Q$.
Analysis 1: Connect $P Q$ and $O M$. By the properties of tangents to a circle, we know $O M \perp P Q$, and the intersection point $E$ of $P Q$ and $O M$ is the midpoint of $P Q$. Let $M(x, y)$, then $|O M|^{2}=x^{2}+y^{2}, |O E|=\frac{|O P|^{2}}{|O M|}=\frac{1}{\sqrt{x^{2}+y^{2}}}$. Thus, the coordinates of point $E$ are $\left(\frac{x}{x^{2}+y^{2}}, \frac{y}{x^{2}+y^{2}}\right)$. Since $\angle P A Q=\frac{\pi}{2}$, we have $|E A| - |E P|$. Also, $|E P|=\sqrt{1-|O E|^{2}}$, so we have $1-|O E|^{2}=|E A|^{2}$, which gives us
$$
1-\frac{1}{x^{2}+y^{2}}=\left(\frac{x}{x^{2}+y^{2}}\right)^{2}+\left(\frac{y}{x^{2}+y^{2}}-\frac{1}{2}\right)^{2} \text {, }
$$

Simplifying, we get $x^{2}+y^{2}+\frac{4}{3} y=\frac{8}{3}$. The above represents a circle with center $\left(0,-\frac{2}{3}\right)$ and radius $\frac{2 \sqrt{7}}{3}$.

Analysis 2: Let the coordinates of $P$ and $Q$ be $\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)$. From the problem, the equations of the tangents at $P$ and $Q$ are
$$
\begin{array}{rlrl}
x_{1} x+y_{1} y & =1, & & \\
x_{1}^{2}+y_{1}^{2} & =1, & & x_{2} x+y_{2} y=1, \\
& & x_{2}^{2}+y_{2}^{2}=1,
\end{array}
$$

Since $P A \perp A Q$, we have $x_{1} x_{2}+\left(y_{1}-\frac{1}{2}\right)\left(y_{2}-\frac{1}{2}\right)=0 \quad$ (5) If (1) and (2) $(x, y)$, we get $y_{1}=\frac{1-x_{1} x}{y}, y_{2}=\frac{1-x_{2} x}{y}$, substituting into (3) and (4), we get
$$
\begin{array}{l}
x_{1}^{2}+\left(\frac{1-x_{1} x}{y}\right)^{2}=1 \Rightarrow\left(x^{2}+y^{2}\right) x_{1}^{2}-2 x_{1} x=y^{2}-1, \\
x_{2}^{2}+\left(\frac{1-x_{2} x}{y}\right)^{2}=1 \Rightarrow\left(x^{2}+y^{2}\right) x_{2}^{2}-2 x_{2} x=y^{2}-1,
\end{array}
$$

Combining the above two equations, we get $\left(x^{2}+y^{2}\right) x_{1}^{2}-\left(x^{2}+y^{2}\right) x_{2}^{2}=2 x\left(x_{1}-x_{2}\right)$. Since $x_{1} \neq x_{2}$, we have $\left(x^{2}+y^{2}\right)\left(x_{1}+x_{2}\right)=2 x$, i.e., $x_{1}+x_{2}=\frac{2 x}{x^{2}+y^{2}}$. Similarly, we get $y_{1}+y_{2}=\frac{2 y}{x^{2}+y^{2}}$. Therefore,
$$
\left(x_{1}+x_{2}\right)^{2}+\left(y_{1}+y_{2}\right)^{2}=\frac{4}{x^{2}+y^{2}} \Rightarrow x_{1} x_{2}+\overline{y_{1}} y_{2}+1=\frac{2}{x^{2}+y^{2}},
$$

From (5), we get $x_{1} x_{2}+y_{1} y_{2}-\frac{1}{2}\left(y_{1}+y_{2}\right)+\frac{1}{4}=0$. Combining the above, we get $\frac{2}{x^{2}+y^{2}}-\frac{y}{x^{2}+y^{2}}=\frac{3}{4}$, which gives us $x^{2}+y^{2}+\frac{4}{3} y=\frac{8}{3}$. The above represents a circle with center $\left(0,-\frac{2}{3}\right)$ and radius $\frac{2 \sqrt{7}}{3}$.