Let $AO \cap \text{ arc}(BD) =K$ and $ON \perp AB$ such that $L$ lies on $AB$....Also let $O$ be center of $\Gamma$ If radius of $\Gamma=r$ then,$$OL^2=(1-r)^2-r^2=\boxed{1-2r} \text{ and } OL^2=(1+r)^2-(1-r)^2=\boxed{4r} \implies 4r=1-2r \implies \boxed{r=\frac{1}{6}}$$

sorry for a long bump but it seems mr.parmenides51 has copied the question wrongly. It should be touches the arc BD internally.

Very similar to CRMO region 2 p1 and region 3 p1. Here, the equation is $(r+1)^2-(r-1)^2=(r-1)^2-r^2\implies r=\frac{1}{6}$