The barycenter is the centroid, or the center of gravity. WLOG let the radius of the circle be 1. For each point $P$ give it a score of $1-OP^2$ where $O$ is the center of the circle. Then we claim that the total score of all points in the circle will always strictly increase. Note that there are only finitely many barycenters (at most $2^n$), so the process will terminate. Now we show the points will always increase. Let $C$ be the barycenter of the points in the circle, and $c$ is the vector $\vec{OC}$. Assume $c\neq 0$. Let $a_1,...,a_m$ be the points in the circle, where each $a_i$ are vectors with origin $C$. Then $\sum a_i=0$. Then the original score is $m-\sum(a_i+c)^2=m-mc^2-\sum(a_i^2)-2c\cdot \sum a_i=m-mc^2-\sum a_i^2$. Note that the scores of points in the circle is positive while those outside is negative. So the total score of the points $a_i$ after moving the circle to $C$ is $m-\sum a_i^2$. The total score of the points in the circle is that value minus scores of points $a_i$ with negative score, plus scores of points not among the $a_i$ with positive score. Thus the total score is $\geq m-\sum a_i^2>m-mc^2-\sum a_i^2$, the original score. Thus the total score strictly increases and we are done.

But how do we know that the displacement of the centre from $C $ to $O $ does not lead to the addition of a new point inside the circle?