Solution:

We show that the orthocenter, normally called $H$, of the triangle $ABC$, here denoted $J$, has the described property. To this end, let $a$ be the line through $A$ parallel to $BC$, and let the lines $b$ and $c$ be defined analogously. No two of the three lines $a$, $b$ and $c$ are parallel, and consequently the intersection point $A'$ of $b$ with $c$ exists, as well as the two analogously defined points $B'$ and $C'$.

The quadrilateral $A' C J B$ has right angles at $B$ and $C$ and is therefore a cyclic quadrilateral. Since reflection at the midpoint of the segment $\overline{BC}$ maps the point $A$ to $A'$ and interchanges $B$ and $C$, it carries $\omega$ into the circumcircle of the cyclic quadrilateral just found. Hence $A_{2}$ lies on this circle, and $A' A_{2}$ is parallel to $AX$.

Let $S$ denote the centroid of the triangle $ABC$. The homothety $\sigma$ with center $S$ and factor $-2$ maps $A$ to $A'$; let the image point of $X$ be called $X'$. Then the two lines $AX$ and $A' X''$ are parallel, and hence $X'$ lies on the line $A' A_{2}$. Similar arguments can also be carried out with $B$ and $C$ in place of $A$, and altogether we learn: the three lines $A' A_{2}$, $B' B_{2}$ and $C' C_{2}$ intersect at $X'$.

If $X' = J$, then the points $A_{2}$, $B_{2}$ and $C_{2}$ also coincide with $J$ and the claim is trivial. From now on, therefore, let $X' \neq J$.

Since $\angle A' C J = 90^{\circ}$, by the theorem of Thales the segment $\overline{A' J}$ is a diameter of the circumcircle of the quadrilateral $A' C J B$. Again by the theorem of Thales we therefore have $\angle J A_{2} A' = 90^{\circ}$ and consequently also $\angle J A_{2} X' = 90^{\circ}$. The point $A_{2}$ therefore lies, once again by the theorem of Thales, on the circle with diameter $J X'$. For analogous reasons the two points $B_{2}$ and $C_{2}$ also lie on this circle, and in particular we have now found a circle on which all four of the points $A_{2}$, $B_{2}$, $C_{2}$ and $J$ lie.