6. Given that $A B$ is the diameter of a semicircle $O$ with a radius of 4, $C$ is the midpoint of the semicircle arc, $E$ is the midpoint of $B C$, a small circle is tangent to $B C$ at $E$ and tangent to $\overparen{B C}$ at $F$, then the length of the tangent $A T$ drawn from $A$ to the small circle is ( ).
(A) $2+\sqrt{2}$
(B) $1+\sqrt{5}$
(C) $2+\sqrt{3}$
(D) $7-2 \sqrt{3}$
6. A.

Connect $A C$, then $A C \perp B C$, connect $A E$ intersecting the smaller circle at $D$, connect $O F$ which must pass through $E$, and connect $D F$. In the right triangle $\triangle A C E$,
$$
\sqrt{A C^{2}+C E^{2}}=\sqrt{(2 \sqrt{2})^{2}+(\sqrt{2})^{2}}=\sqrt{10}.
$$
$A E=$
It is also easy to prove that right triangles $\triangle E D F \sim \triangle A C E$, so $\frac{E D}{A C}=\frac{E F}{A E}$. Therefore, $E D=\frac{A C(O F-O E)}{A E}=\frac{2 \sqrt{2}(2-\sqrt{2})}{\sqrt{10}}$, hence $A D=A E+E D$,
which means $A T=\sqrt{A E \cdot A J}=\sqrt{5+4 \sqrt{2}}=2+\sqrt{2}$.