jmc

Let $X$ be the intersection of the circles with centers $B$ and $E$, and $Y$ be the intersection of the circles with centers $C$ and $E$. Since the radius of $B$ is $3$, $AX=4$. Assume $AE$ = $p$. Then $EX$ and $EY$ are radii of circle $E$ and have length $4+p$. $AC=8$, and angle $CAE=60$ degrees because we are given that triangle $T$ is equilateral. Using the Law of Cosines on triangle $CAE$, we obtain $(6+p{)}^2=p^2+64-2(8)(p)\cos 60$. The $2$ and the $\cos 60$ terms cancel out: $p^2+12p+36=p^2+64-8p$ $12p+36=64-8p$ $p=\frac{28}{20}=\frac{7}{5}$. The radius of circle $E$ is $4+\frac{7}{5}=\frac{27}{5}$, so the answer is $27+5=\boxed{32}$.