   too complicated to be written here. Click on the link to download a text file.  X(523), X(3448), X(6328) A'B'C' : cevian triangle of X(3448) PaPbPc : anticevian triangle of X(523) in/excenters of PaPbPc Q = X(3448)-Ceva conjugate of X(523), perspector of A'B'C' and PaPbPc    K878 is the only pK with pole X(115) which is a circular cubic. Its singular focus F = X(14695) is the reflection of its pivot X(3448) about X(10279), the circum-center of PaPbPc. K878 is also an isogonal circular pK with respect to PaPbPc and its pivot is then X(523). It is actually the only isogonal pK with respect to PaPbPc which is also a circum-cubic of ABC. See a generalization below.  The orthic line of K878 passes through X(140), X(523), X(1116), X(10279) hence the polar conic of each point on this line is a rectangular hyperbola. In particular, the polar conic (H) of X(523) passes through X(30), X(523) and the in/excenters Jo, Ja, Jb, Jc of PaPbPc. See the coordinates below. These four points are the centers of anallagmaty of the cubic and the tangents are parallel to the orthic line and perpendicular to the Euler line of ABC. These tangents are also parallel to the lines APa, BPb, CPc. The real asymptote is the line X(523)X(3448) hence X(3448) is the point where the cubic meets its asymptote. It is the antipode of the singular focus F on the circum-circle (C) of PaPbPc.    Let A1 = Sqrt [a^4 (– a^2 + 3 b^2 + 3 c^2) – 3 a^2 b^2 c^2 + (b^4 + c^4)(– 3 a^2 + b^2 + c^2)], B1 and C1 cyclically. The incenter Jo of PaPbPc is : (b^2 – c^2) [– A1 (b^2 – c^2) + B1 (c^2 – a^2) + C1 (a^2 – b^2)] : (c^2 – a^2) [A1 (b^2 – c^2) – B1 (c^2 – a^2) + C1 (a^2 – b^2)] : (a^2 – b^2) [A1 (b^2 – c^2) + B1 (c^2 – a^2) – C1 (a^2 – b^2)], with SEARCH = 2.74926319443787. The excenters Ja, Jb, Jc are obtained from Jo by successively replacing A1, B1, C1 with their opposites. The corresponding SEARCH numbers are : –16.4866509137355; 6.47020826139626; 3.84311030746983.    Generalization Let P be a point with anticevian triangle PaPbPc. Let K(P, Q) be the pivotal pK with pivot Q which is self-isogonal with respect to PaPbPc. K(P,Q) is generally not a circum-cubic of ABC unless Q = P. In this case K(P, P) = K(P) is : • the isogonal pK with pivot P with respect to PaPbPc, • the pK in ABC with pole Ω = P^2 (barycentric square), pivot S the isogonal conjugate of P with respect to PaPbPc. Examples • with P = X(1), PaPbPc is the excentral triangle, K(P) is K414, the only cubic of this type which is an isogonal pK in both triangles. • with P = X(2), PaPbPc is the antimedial triangle, K(P) is the Lucas cubic K007 which is an isotomic pK in ABC, an isogonal pK in PaPbPc since it is the Thomson cubic for this latter triangle. • with P = X(6), PaPbPc is the tangential triangle, K(P) is K174 = pK(X32, X22). Circular cubics K(P) is circular if and only if P lies on the line at infinity. In this case, the pole Ω lies on the Steiner inellipse, the pivot S lies on the circle C(H, 2R). Obviously, K878 is obtained for P = X(523).  