Let $M,N,L$ be the midpoints of the edges $BC,CD,DB.$ Denote $\Omega \equiv \odot(BCD)$ and $\omega \equiv \odot(MNL).$ Medians $AG_A,BG_B,CG_C,DG_D$ concur at $G$ such that $\overline{GG_A}:\overline{GA}=-1:3,$ etc. Bimedians $NB',LC',MD'$ bisect each other at $G.$ Assume that $A,G,G_B,G_C,G_D$ lie on a same sphere $\mathcal{S}_1.$ If $P \in AG$ such that $\overline{GA}:\overline{GP}=-1:3,$ then $B,C,D,G,P$ lie on the homothetic sphere of $\mathcal{S}_1$ under homothety $(G,-3).$ Thus $p(G_A,\Omega)=\overline{G_AP} \cdot \overline{G_AG}=-\frac{_1}{^3}GA \cdot \frac{_8}{^3}GA=-\frac{_8}{^9}GA^2.$
Assume that $A,B',C',D',G$ lie on the same sphere $\mathcal{S}_2.$ If $Q$ is the reflection of $A$ about $G,$ then $Q,N,L,M,G$ lie on the reflection of $\mathcal{S}_2$ about $G.$ Therefore, $p(G_A,\omega)=\overline{G_AG} \cdot \overline{G_AQ}=-\frac{_1}{^3}GA \cdot \frac{_2}{^3}GA=-\frac{_2}{^9}GA^2.$ But clearly $p(G_A,\omega)=\frac{_1}{^4}p(G_A,\Omega),$ thus $p(G_A,\Omega)=-\frac{_8}{^9}GA^2.$ Consequently, $A,G,G_B,G_C,G_D$ lie on a same sphere $\Longleftrightarrow$ $A,B',C',D',G$ lie on a same sphere.