Let $ABC$ be a triangle, Let $(W)$ be the circle through $X(3),X(110)$ and $X(4240)$ in $ETC$. I found 7 points also lie on $(W)$ as follows:
Let the Euler line of ABC meets the sidelines $BC,CA,AB$ at $A_0,B_0,C_0$. Six points $X(3), X(110)$ of three triangles $AB_0C_0, BA_0C_0, CB_0A_0$ also lie on $(W)$. The circumcenter $X(3)$ of Paralogic triangle whose perpectrix is the Euler line of $ABC$ also lie on $(W)$. So $10$ points lie on this circle.
Center of this circle is midpoint of $X(3)$ of $ABC$ and $O'$
Let the Euler line of ABC meets the sidelines $BC,CA,AB$ at $A_0,B_0,C_0$. Six points $X(3), X(110)$ of three triangles $AB_0C_0, BA_0C_0, CB_0A_0$ also lie on $(W)$. The circumcenter $X(3)$ of Paralogic triangle whose perpectrix is the Euler line of $ABC$ also lie on $(W)$. So $10$ points lie on this circle.
Center of this circle is midpoint of $X(3)$ of $ABC$ and $O'$
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