Denote by $(I,r)$ and $(O,R)$ the fixed incircle and circumcircle of $\triangle ABC.$ $\triangle DEF$ and $\triangle A_0B_0C_0$ are the medial and antimedial triangle of $\triangle ABC$ with incenters $S,N,$ the Spieker point and Nagel point of $\triangle ABC,$ respectively. It's well-known that $S$ is the radical center of the three excircles of $\triangle ABC$ (see Hard geometry problem), further, $N$ is the reflection of $I$ about $S.$ Since the incircle $(N,2r)$ and 9 point circle $(O,R)$ of $\triangle A_0B_0B_0$ are internally tangent, we obtain $ON=R-2r.$ So if $M$ is the midpoint of $\overline{IO},$ we have $MS=\frac{_1}{^2}ON=\frac{_1}{^2}R-r=\text{const}$ $\Longrightarrow$ Locus of $S$ is then a circle with center $M$ and radius $\frac{_1}{^2}R-r.$

in fact if $O,I$ are fixed then $H$ which is the orthocenter lies on a fixed circle which center is the reflection of $O$ wrt $I$.this was in a crux magazine ... but unfortunately i don't remember the vol and this quickly solve the problem as the radical center is $I'$ the incenter of the median triangle collinear with $G,I$ and $G$ collinear with $O,H$.