Circles $$A$$, $$B$$ and $$C$$ are externally tangent to each other and internally tangent to circle $$D$$. Circles $$A$$ and $$B$$ are congruent. Circle $$C$$ has radius $$1$$ unit and passes through the centre of circle $$D$$. Then the radius of circle $$B$$ is

Let the radius of circles $$A$$ and $$B$$ be $$r$$, and let the relevant horizontal distance between the centres be $$x$$. From the tangency geometry, $$x^2=4(1-r)$$ and also $$x^2+2x=2r$$. Eliminating $$x$$ gives $$3x^2+4x-4=0$$, so the positive solution is $$x=\frac{2}{3}$$. Substitution gives $$r=\frac{8}{9}=0.8888888889$$.