smc

Let $AG=A{A}^{\prime }=a$ and $BG=B{B}^{\prime }=b$. Then, we know that the diameter $d$ of the circle equals $AG+BG=a+b$. Thus, because the circle's diameter does not change, $a+b$ is constant. Because $A^{\prime }{O}^{\prime }=O^{\prime }{B}^{\prime }$ and $AO=OB$, $\overline{O{O}^{\prime }}\parallel \overline{A{A}^{\prime }}$. Thus, $\overline{O{O}^{\prime }}\perp \overline{AB}$, and so $O{O}^{\prime }$ is the distance from $O^{\prime }$ to $\overline{AB}$. Let $P$ be some point which moves along $\overline{A^{\prime }{B}^{\prime }}$. Because $A^{\prime }{B}^{\prime }$ is a line segment, as $P$ moves from $A^{\prime }$ to $B^{\prime }$, its distance from $\overline{AB}$ will vary linearly with how much it has travelled along $\overline{A^{\prime }{B}^{\prime }}$. Thus, when it is halfway along $\overline{A^{\prime }{B}^{\prime }}$ (in other words, when $P=O^{\prime }$) its distance from $\overline{AB}$ will be the arithmetic mean of its distance from $\overline{AB}$ at $A^{\prime }$ (namely, $A^{\prime }A=a$) and its distance from $\overline{AB}$ at $B^{\prime }$ (namely, $B^{\prime }B=b$). Thus, $O^{\prime }O=\frac{a+b}{2}$. Because $a+b=d$, a constant, $\frac{a+b}{2}$ is a constant as well. Thus, $O{O}^{\prime }$ is the same regardless of the position of $G$. Furthermore, from our work in paragraph 2, we know that $O^{\prime }$ must lie on the line perpendicular to $\overline{AB}$ through point $O$. Therefore, because $O^{\prime }$ is a fixed distance from a fixed point on a fixed line, and it will not suddenly "jump across" to the other side of $\overline{AB}$, we can say with confidence that point $O^{\prime }$ $\boxed{\text{remains stationary}}$.