Circless

Let the radius of the cycle and semicircle be $r$ and $r_1$ and two heights $a$ and $b$ as shown in the figure. The radius of the quarter circle is 2. Then from the large right triangle and using Pythagorean theorem , we have:

$\begin{aligned} (2+r_1)^2 - (2-r_1)^2 & = 2^2 \\ 8 r_1 & = 4 \\ \implies r_1 & = \frac 12 \end{aligned}$

From the top right triangle:

$\begin{aligned} (2+r)^2 - (2-r)^2 & = a^2 \\ 8 r & = a^2 \\ \implies a & = 2\sqrt {2r} \end{aligned}$

From the small right triangle:

$\begin{aligned} \left(\frac 12+r\right)^2 - \left(\frac 12+r\right)^2 & = b^2 \\ 2r & = b^2 \\ \implies b & = \sqrt {2r} \end{aligned}$

Since $a+b=2$ ,

$\begin{aligned} 2\sqrt{2r} + \sqrt{2r} & = 2 \\ 3\sqrt{2r} & = 2 \\ 9(2r) & = 4 \\ \implies r & = \boxed{\dfrac 29} \end{aligned}$