Example 13 Consider the equation $m^{2}+(k-4) m$ $+k=0$ with respect to $m$. Let the largest integer root of this equation be the diameter of $\odot O$. Let $P$ be a point outside $\odot O$, and draw the tangent $P A$ and the secant $P B C$ through $P$, as shown in Figure 1, with $A$ being the point of tangency. It is found that $P A, P B, P C$ are all integers, and $P B, B C$ are not composite numbers. Find the lengths of $P A, P B, P C$.
Solution: Let the two roots of the equation be $m_{1}$ and $m_{2}$, then
$$
\left\{\begin{array}{l}
m_{1}+m_{2}=4-k, \\
m_{1} m_{2}=k .
\end{array}\right.
$$

Let $P A=x, P B=y, B C=z$, then $x, y, z$ are all positive integers.
By the secant-tangent theorem, we have
$$
P A^{2}=P B \cdot P C=P B(P C+3 C),
$$

which means
$$
x^{2}=y^{2}+y \Rightarrow(x+y)(x-y)=y z \text {. }
$$

Eliminating $k$ from (1) and the equation, we get
$$
m_{1} m_{2}=4-m_{1}-m_{2} \text {. }
$$

Rearranging and factoring, we get
$$
\left(m_{1}+1\right)\left(m_{2}+1\right)=5 \text {. }
$$

Since the diameter of $\odot O$ is the largest integer root of the equation, it is not difficult to find that the maximum integer root $m=4$. Thus, $z=B C \leqslant 4$.
Since the positive integer $z$ is not a composite number, hence $z=3,2,1$.
When $z=3$, $(x+y)(x-y)=3 y$, we have
$$
\left\{\begin{array}{l}
x+y=3, \\
x-y=y ;
\end{array} \left\lvert\, \begin{array}{l}
x+y=y, \\
x-y=3 ;
\end{array}\left\{\begin{array}{l}
x+y=3 y, \\
x-y=1 .
\end{array}\right.\right.\right.
$$

The solution that fits the problem is $x=2, y=1$.
When $z=1$ and $z=2$, there are no solutions that fit the problem,
Therefore, $P A=x=2, P B=y=1, P C=y+$ $z=4$.