6. As shown in Figure 3, given two circles $\odot O_{1}$ and $\odot O_{2}$ with radii 1 and 2 respectively, which are externally tangent to each other, line $A B$ is a common external tangent to the two circles, touching $\odot O_{1}$ and $\odot O_{2}$ at points $A$ and $B$ respectively. $A C$ and $B D$ are diameters of the two circles, $C E$ is tangent to $\odot O_{2}$ at point $E$, and $D F$ is tangent to $\bigcirc O_{1}$ at point $F$. Then the value of $|C E-D F|$ is $(\quad)$.
(A) $\sqrt{2}$
(B) $\sqrt{3}$
(C) 2
(D) $\sqrt{5}$
6. C.

As shown in Figure 6, let the point of tangency of the two circles be $M$. Draw the common internal tangent $MN$ through $M$, intersecting $AB$ at point $N$. Connect $AM$, $BM$, and $CM$. It is easy to see that
$$
\begin{aligned}
& \angle AMB \\
= & \angle AMC=90^{\circ} .
\end{aligned}
$$

Therefore, points $B$, $M$, and $C$ are collinear.
In the right triangle $\triangle ABC$, by the projection theorem, we have
$$
\begin{array}{l}
AC^{2}=CM \cdot CB \\
\Rightarrow AC^{2}=CE^{2} \\
\Rightarrow CE=AC=2 .
\end{array}
$$

Similarly, $DF=BD=4$.
Thus, $|CE - DF|=2$.