[Antreas P. Hatzipolakis]:

As for our communication of Anopolis #816 and others, I obtained the following results.

[1] Given a triangle ABC, its Euler line L and its orthic triangle DEF, let D’, E’, F’ be the reflections of D, E, F in L respectively, the circumcircles of D’BC, E’CA, F’AB concur in X(115).

[2] Given a triangle ABC, its Euler line L and its orthic triangle DEF, let D’, E’, F’ be the reflections of D, E, F in L respectively, the circumcircles of AE’F’, BF’D’, CD’E’ concur in X(935).

[3] Given a triangle ABC, its Euler line L and its orthic triangle DEF, let A’, B’, C’ be the reflections of A, B, C in L respectively, the circumcircles of A’EF, B’FD, C’DE concur in X(112).

[4] Given a triangle ABC, its Euler line L and its orthic triangle DEF, let A’, B’, C’ be the reflections of A, B, C in L respectively, the circumcircles of DB’C’, EC’A’, FA’B’ concur in a point P.

Bricentrics of P: {(b^2-c^2)^2(b^2+c^2-2a^2)(b^4+c^4-a^4-b^2c^2); ; }.

I replaced Euler line by a general line {l,m,n}. Mathematica produced the factor of (l(a^4-(b^2-c^2)^2)+m(b^4-(c^2-a^2)^2)+n(c^4-(a^2-b^2)^2) for the difference between the radical center and the radius of the circumcircle. So if the line passes the orthocenter of ABC, 3 circles concur in a point.