Example 1. The vertex of the parabola is at the origin, and the focus F is the center of the circle given by $x^{2}+y^{2}-4 x=0$. A line passing through point $F$ with a slope of 2 intersects the parabola at points $A$ and $D$, and intersects the circle at points $B$ and $C$.
Find $|A B|+|C D|$.
$$
\begin{array}{l}
\text { Given } F \text { as the pole, and the positive direction of the } x \text { axis as the polar axis, we establish the polar coordinate system. } \\
\left.\begin{array}{l}
\left.\therefore \text { The parabola equation is: } \rho=\frac{p}{1-\cos \theta}\right\} \Rightarrow \rho= \\
\because F(2,0), \therefore p=4 .
\end{array}\right\} \Rightarrow
\end{array}
$$
$$
\begin{array}{l}
1-\cos \theta \text {. } \\
\text { Also, } \because \operatorname{tg} \theta=2, \therefore \cos \theta=\frac{\sqrt{5}}{5}(0 \leqslant \theta<\pi) \text {. } \\
\because|A B|+|C D|=|A F|+|F D|-|B C| \text {, } \\
|A F|=\rho_{1}, \quad|F D|=\rho_{2}, \\
\therefore \rho_{1}+\rho_{2}=\frac{4}{1-\cos \theta}+\frac{4}{1-\cos (\theta+\pi)} \\
=\frac{8}{1-\cos ^{2} \theta}=\frac{1}{1-\frac{1}{5}}=10 \text {. } \\
\end{array}
$$
$$
\begin{array}{c}
\text { Also, } \because|B C|=2 R=4, \\
\therefore|A B|+|C D|=10-4=6 .
\end{array}
$$