A geometry problem by A Former Brilliant Member

Let $r$ be the radius of circle $E$ .

$AF=\sqrt{10^2-6^2}=8$

$CH=CF=6;AH=4$

Since $\triangle AHO \sim \triangle AFC$ , then, $\dfrac{OH}{FC}=\dfrac{AH}{AF}$ $\implies$ $\dfrac{OH}{6}=\dfrac{4}{8}$ $\implies$ $OH=3$

$\large \therefore GF=6$ and $AG=2$

$EA=2-r$

Since $ED \parallel OH$ , $\triangle ADE \sim \triangle AHO$ . Then,

$\dfrac{ED}{EA}=\dfrac{OH}{OA}$ $\implies$ $\dfrac{r}{2-r}=\dfrac{3}{5}$ $\implies$ $5r=6-3r$ $8r=6$ $\implies$ $r=\dfrac{6}{8}=\dfrac{3}{4}=$ $\boxed{0.75}$