Find the radius of the circle given in the picture below. This is the image of the question. I am not able to get how to find the radius. Please help with that.

This is my try. I can't proceed now after it.

Thanks


Given: $CD=2,\,BD=3,\,AB=\sqrt{11}$.
Let's define $\angle B=\angle DBE,\, CE=d$.
$2R=\frac{\sqrt{11}}{\cos A}$, $BE=\frac{\sqrt{11}}{\sin A}, \angle DBE=\angle DAE$, thus $CBE\sim CAD$
$$\frac{CB}{CA}=\frac{BE}{AD}=\frac{CE}{CD}$$
$$\frac{5}{d+2R}=\frac{\frac{\sqrt{11}}{\sin A}}{AD}=\frac{d}{2}$$
By applying the cosine rule to both $\Delta CBA$, $\Delta CDA$ we get
$$\begin{cases}
11=25+(2R+d)^2-2\cdot 5\cdot (2R+d)\cos C,\\
AD^2=4+(2R+d)^2-2\cdot 2\cdot (2R+d)\cos C,
\end{cases}$$
By eliminating $\cos C$ we get $$
AD^2=4+(2R+d)^2-\frac{2}{5}\left(25+(2R+d)^2-11\right)=$$
$$\frac{3}{5}(2R+d)^2-\frac{12}{5}=15\left(\frac{2R+d}{5}\right)^2-\frac{12}{5}=\frac{60}{d^2}-\frac{12}{5}$$
Thus we get $$\begin{cases}
10=d(d+2R)\\
2\sqrt{11}=d\sqrt{\frac{60}{d^2}-\frac{60}{25}}\sin A
\end{cases}$$
$$\begin{cases}
10=d(d+\frac{\sqrt{11}}{\cos A})\\
2\sqrt{11}=\frac{\sqrt{60}}{5}\sqrt{25-d^2}\sin A
\end{cases}$$ hence we get $d$ and then $R$.

