$R=\frac{\left|BD\right|}{2\sin \alpha }$
       $\{\begin{array}{l}|BD{|}^2=3^2+8^2-2\cdot 3\cdot 8\cdot \cos \alpha \\ |BD{|}^2=3^2+5^2+2\cdot 3\cdot 5\cdot \cos \alpha \end{array}$
$$\begin{gathered} \begin{array}{r}\\ -\end{array}\{\begin{array}{l}|BD{|}^2=9+64-48\cos \alpha \\ |BD{|}^2=9+25+30\cos \alpha \end{array} \\ \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \end{gathered}$$
            $0=39-78\cos \alpha$
            $\cos \alpha =\frac{39}{78}=\frac{1}{2}$    $\to$    $\alpha =60^{\circ }$

$|BD{|}^2=9+25+30\cdot \frac{1}{2}=34+15=49$
$\left|BD\right|=7$

$R=\frac{\left|BD\right|}{2\sin \alpha }=\frac{7}{2\sin 60^{\circ }}=\frac{7}{2\cdot \frac{\sqrt{3}}{2}}=\frac{7}{\sqrt{3}}=\frac{7\sqrt{3}}{3}$

Odp.: Promień tego okręgu jest równy $\frac{7\sqrt{3}}{3}$.