6. As shown in the figure, a semicircle with radius

1 has its center at $O$ and diameter $A B$. A moving point $C$ is on the entire
circumference. Extend $A C$ to point $D$ such that $\frac{C D}{C B}=k(k>0, k$ 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) $\pi \cdot \sqrt{1+k^{2}}$
6. (D).

As shown in the figure, let $\angle C A B=\theta\left(0 \leqslant \theta \leqslant \frac{\pi}{2}\right)$, then $\angle C O B=2 \theta, C B=2 \sin \theta, C D=2 k \sin \theta, C$ point's coordinates are $(\cos 2 \theta, \sin 2 \theta)$. Draw $C E / / A B$, then $\angle D C E = \angle C A B = \theta$. Let the coordinates of $D$ point be $(x, y)$, then
$$
\left\{\begin{array}{l}
x=\cos 2 \theta+2 k \sin \theta \cos \theta=\cos 2 \theta+k \sin 2 \theta, \\
y=\sin 2 \theta+2 k \sin \theta \sin \theta=\sin 2 \theta+k(1-\cos 2 \theta) .
\end{array}\right.
$$

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

Then, by $\cos ^{2} 2 \theta+\sin ^{2} 2 \theta=1$ we get
$$
\begin{array}{l}
(y+k x-k)^{2}+\left(x-k y+k^{2}\right)^{2} \\
=\left(1+k^{2}\right)^{2},
\end{array}
$$

Expanding and simplifying, we get $x^{2}+y^{2}-2 k y=1$, i.e., $x^{2}+(y-k)^{2}=\left(\sqrt{1+k^{2}}\right)^{2}$,
and also $k x+y \geqslant k$. Therefore, the trajectory of $D$ point is the upper half of a circle with the line segment connecting points $F(-1,2 k)$ and $B(1,0)$ as its diameter, with the center at $G(\hat{u}, k)$, where $\hat{u} = \sqrt{1+k^{2}}$, so its trajectory length is $\pi \sqrt{1+k^{2}}$.