For the sake of generality, let $\mathcal{O} \equiv (O,R)$ be the circumsphere of the tetrahedron $ABCD$ and $ \mathcal{S} \equiv (S,R')$ be the circumsphere of the tetrahedron whose vertices are the centroids $A',$ $B',$ $C',$ $D'$ of its faces against $A,B,C,D.$ Segments $AA',$ $ BB',$ $ CC',$ $ CC',$ $DD'$ concur at the centroid $G$ of $ABCD,$ such that $G$ divides them in the same ratio $1:3$ $\Longrightarrow$ Tetrahedra $A'B'C'D'$ and $ABCD$ are homothetic through the homothety with center $G$ and coefficient $-\frac{_1}{^3}.$ Therefore, $R'=\frac{_1}{^3}R$ and $OS=\frac{_4}{^3}OG.$ Let $a,b,c,d,e,f$ be the edges of $ABCD.$ By Leibniz theorem for the circumcenter $O$ of $ABCD,$ we have $OG^2=\frac{_1}{^4}(OA^2+OB^2+OC^2+OD^2)-\frac{_1}{^{16}}(a^2+b^2+c^2+d^2+e^2+f^2)$ $OG^2=R^2-\frac{_1}{^{16}}(a^2+b^2+c^2+d^2+e^2+f^2)$ $\Longrightarrow OS^2=\frac{_{16}}{^9}R^2-\frac{_1}{^9}(a^2+b^2+c^2+d^2+e^2+f^2)$