smc

Let $O_i$ be the center of circle $i$ for all $i\in \{A,B,C,D\}$ and let $E$ be the tangent point of $B,C$. Since the radius of $D$ is the diameter of $A$, the radius of $D$ is $2$. Let the radius of $B,C$ be $r$ and let $O_{D}E=x$. If we connect $O_A,O_B,O_C$, we get an isosceles triangle with lengths $1+r,2r$. Then right triangle $O_D{O}_B{O}_E$ has legs $r,x$ and hypotenuse $2-r$. Solving for $x$, we get $x^2=(2-r{)}^2-r^2⟹x=\sqrt{4-4r}$. Also, right triangle $O_A{O}_B{O}_E$ has legs $r,1+x$, and hypotenuse $1+r$. Solving, \begin{eqnarray} r^2 + (1+\sqrt{4-4r})^2 &=& (1+r)^2\\ 1+4-4r+2\sqrt{4-4r}&=& 2r + 1\\ 1-r &=& \left(\frac{6r-4}{4}\right)^2\\ \frac{9}{4}r^2-2r&=& 0\\ r &=& \frac 89 \end{eqnarray} So the answer is $\boxed{\frac{8}{9}}$.