This familiar triangle again!

Typically, I am using the dirty Trigonometry to solve this pure and beautiful Geometry problem. Then again, I'm a Chinese. And whether it is a white cat or black cat, so long as it catches mice, it is a good cat. Specifically, I am using the fact that the line joining a vertex of a triangle to its incenter bisects the vertex angle.

Let the radius of the red circle be $r$ , then the radius of the blue circle is $2r$ and $AD = 6r$ . Let $\angle BAD = \alpha$ . Consider the red circle and $AD$ :

$\begin{aligned} r \cot \frac \alpha 2 + r & = AD = 6r \\ \cot \frac \alpha 2 + 1 & = 6 \\ \implies \tan \frac \alpha 2 & = \frac 15 \\ \tan \alpha & = \frac {2\cdot \frac 15}{1-\frac 1{25}} = \frac 5{12} & \small \blue{\implies BD = 2.5r, \ AB = 6.5r} \end{aligned}$

Similarly, for the green circle and $AD$ :

$\begin{aligned} 3 \cot \frac \beta 2 + 3 & = 6r \\ \tan \frac \beta 2 & = \frac 1{2r-1} \end{aligned}$

The blue circle and $AB$ :

$\begin{aligned} 2r \cot \frac {\alpha+\beta}2 + 2r \cot \frac {90^\circ - \alpha}2 & = 6.5 r \\ 4 \cdot \frac {1 - \frac 1{5(2r-1)}}{\frac 15 + \frac 1{2r-1}} + 4 \cdot \frac {1+\frac 15}{1-\frac 15} & = 13 \\ \frac {20r-12}{r+2} + 6 & = 13 \\ 20r - 12 & = 7r+14 \\ \implies r & = 2 & \small \blue{\implies \tan \frac \beta 2 = \frac 13, \ \tan \beta = \frac 34, \ DC = 9} \end{aligned}$

Then the circumradius $R$ of $\triangle ABC$ is given by:

$\begin{aligned} 2R & = \frac {BC}{\sin \angle BAC} = \frac {5+9}{\sin (\alpha + \beta)} = \frac {14 \cdot 65}{56} \\ \implies R & = \frac {65}8 = \boxed{8.125} \end{aligned}$