Geometry

Let the center of the small circle be O, and the point where circle O is tangent to the circle with diameter AC be D, and let the radius of circle O be r. Lastly, let the midpoints of AB and BC be points M and N, respectively.

Since B is the center of a semicircle, BD must also be the same length as AB (because they are the radius) = 2.

MO = 1 + r, and MB = 1. OMB forms a right triangle with angel OBM = 90 degrees, since O lies on the perpendicular bisector of AC.

This means that OB, by the pythagorean theorem, has length $\sqrt{(1+r)^2-1}=\sqrt{r^2+2r}$.

Additionally, DO = r, and lines on the same line as OB. Recalling that DB = 2, we can set an equation: DO + OB = 2:

$r + \sqrt{r^2+2r}=2$

$\sqrt{r^2+2r}=2-r$

$r^2+2r=(2-r)^2=r^2-4r+4$

$2r = -4r+4$

$6r=4$

$r={2\over3}$

Let the center of the smaller semi-circle be $X$ and the center of the small circle be $O$. 
$\triangle BOX$ is a right angle triangle with right angle at $B$. So:

$BO^2+BX^2=OX^2\\ (2-r)^2+1^2=(1+r)^2\\ 6r=4\\ \boxed{r=\displaystyle\frac{2}{3}}$