Let $M$ be the midpoint of chord $A B$ and let $O$ be the center of $\Omega$. Since $A M=B M=3$, Pythagoras on triangle $A M O$ gives $O M=4$. Now let $\omega$ be centered at $P$ and say that $\omega$ and $\Omega$ are tangent at $Q$. Because the diameter of $\omega$ exceeds 1, points $P$ and $Q$ lie on the same side of $A B$. By tangency, $O, P$, and $Q$ are collinear, so that $O P=O Q-P Q=4$. Let $H$ be the orthogonal projection of $P$ onto $O M$; then $O H=O M-H M=O M-P T=3$. Pythagoras on $O H P$ gives $H P^{2}=7$. Finally, $A T \cdot B T=A M^{2}-M T^{2}=A M^{2}-H P^{2}=9-7=2$