jmc

Let $E,H$, and $F$ be the centers of circles $A,B$, and $D$, respectively, and let $G$ be the point of tangency of circles $B$ and $C$. Let $x=FG$ and $y=GH$. Since the center of circle $D$ lies on circle $A$ and the circles have a common point of tangency, the radius of circle $D$ is $2$, which is the diameter of circle $A$. Applying the Pythagorean Theorem to right triangles $EGH$ and $FGH$ gives $$(1+y{)}^2=(1+x{)}^2+y^2\text{and}(2-y{)}^2=x^2+y^2,$$ from which it follows that $$y=x+\frac{x^2}{2}\text{and}y=1-\frac{x^2}{4}.$$ The solutions of this system are $(x,y)=(2/3,8/9)$ and $(x,y)=(-2,0)$. The radius of circle $B$ is the positive solution for $y$, which is $\boxed{\frac{8}{9}}$.