Assume the radius of the $k$-th shaded circle is $r_{k}$. Then, by Descartes' theorem, we have
$$
\left(2-1+\frac{1}{r_{k}}+\frac{1}{r_{k+1}}\right)^{2}=2\left(1+4+\frac{1}{r_{k}^{2}}+\frac{1}{r_{k+1}^{2}}\right) \text {. }
$$

Similarly,
$$
\left(2-1+\frac{1}{r_{k}}+\frac{1}{r_{k-1}}\right)^{2}=2\left(1+4+\frac{1}{r_{k}^{2}}+\frac{1}{r_{k-1}^{2}}\right) \text {. }
$$

Subtracting the two equations, we get
$$
\begin{array}{l}
2\left(\frac{1}{r_{k+1}}-\frac{1}{r_{k-1}}\right)\left(\frac{1}{r_{k+1}}+\frac{1}{r_{k-1}}\right) \\
\quad=\left(\frac{1}{r_{k+1}}-\frac{1}{r_{k-1}}\right)\left(2+\frac{2}{r_{k}}+\frac{1}{r_{k-1}}+\frac{1}{r_{k+1}}\right) \\
\quad \Rightarrow \frac{1}{r_{k+1}}+\frac{1}{r_{k-1}}-\frac{2}{r_{k}}=2 \\
\quad \Rightarrow\left(\frac{1}{r_{k+1}}-\frac{1}{r_{k}}\right)-\left(\frac{1}{r_{k}}-\frac{1}{r_{k-1}}\right)=2 .
\end{array}
$$

Summing up, we get
$\frac{1}{r_{k}}=k^{2}-k+C$ (where $C$ is a constant to be determined).
Therefore, $\frac{1}{r_{1}}=C, \frac{1}{r_{2}}=2+C$.
Since the large circle, the small semicircle, and the first and second shaded circles are tangent, then
$$
\begin{array}{l}
(-1+2+C+2+C)^{2} \\
=2\left[1+4+C^{2}+(C+2)^{2}\right] \\
\Rightarrow C=\frac{9}{4} \Rightarrow r_{k}=\frac{4}{4 k^{2}-4 k+9} .
\end{array}
$$
Example 3 As shown in Figure 5, let the radius of the largest semicircle be 1, and the radius of the smaller semicircle be $\frac{1}{2}$. Find the radius of the $k$-th shaded circle.