$26 \cdot 11$ As shown in the figure, a semicircle with radius 1 is centered at 0, with $A B$ as its diameter. A moving point $C$ travels along the entire circumference. Extend $A C$ to point $D$ such that $\frac{C D}{C B}=k$ (where $k>0, k$ is a constant). Then the length of the curve traced by point $D$ is
(A) $(1+k) \cdot \pi$.
(B) $\pi \cdot \sqrt{1+k}$.
(C) $\left(1+k^{2}\right) \cdot \pi$.
(D) $\pi \cdot \sqrt{1+k^{2}}$.
[Solution] As shown in the figure, let $\angle CAB = \theta \left(0 \leqslant \theta \leqslant \frac{\pi}{2}\right)$. Then
$$
\angle COB = 2\theta, \quad CB = 2 \sin \theta, \quad CD = 2k \sin \theta.
$$
The coordinates of point $C$ are $(\cos 2\theta, \sin 2\theta)$.
Draw $CE \parallel AB$ through point $C$, then $\angle DCE = \angle CAB = \theta$.
Let the coordinates of point $D$ be $(x, y)$, then
$$
\begin{array}{l}
\left\{\begin{array}{l}
x = \cos 2\theta + 2k \sin \theta \cos \theta, \\
y = \sin 2\theta + 2k \sin \theta \sin \theta,
\end{array}\right. \\
\left\{\begin{array}{l}
x = \cos 2\theta + k \sin 2\theta \\
y = \sin 2\theta + k(1 - \cos 2\theta).
\end{array}\right.
\end{array}
$$

Solving, we get $\sin 2\theta = \frac{y + kx - k}{1 + k^2}$
(thereby $kx + y \geqslant k$),
$$
\cos 2\theta = \frac{x - ky + k^2}{1 + k^2}.
$$

Then, by $\cos^2 2\theta + \sin^2 2\theta = 1$,
we have
$$
(y + kx - k)^2 + (x - ky + k^2)^2 = (1 + k^2)^2,
$$

which simplifies to
$$
x^2 + (y - k)^2 = (\sqrt{1 + k^2})^2.
$$

And we also have $kx + y \geqslant k$.
From this, we know: The trajectory of point $D$ is the upper half of a circle with the line segment connecting points $F(-1, 2k)$ and $B(1, 0)$ as its diameter, with the center at $G(0, k)$ and radius $\sqrt{1 + k^2}$. Therefore, the length of its trajectory is $\pi \sqrt{1 + k^2}$.
Thus, the answer is $(D)$.