too complicated to be written here. Click on the link to download a text file. X(6), X(53), X(216), X(393), X(1609), X(2165), X(14576), X(41523), X(41524), X(41525), X(41536) infinite points of K1188 = pK(X3767, X264) and pK(X5, P5) points of pK(X2207, X53), pK(X32, P32), pK(Ω6, X6) on the circumcircle P5, P32, Ω6 are X(41757), X(41758), X(41759) in ETC (2021-03-09)
 Every pivotal cubic pK with pivot H and pole Ω is an isogonal pK with respect to the orthic triangle, its pivot being the tangential of H i.e. the Ω-isoconjugate of H. When Ω lies on K627, this cubic is an isogonal pK with respect to a second triangle and the cubic is said to be a bi-isogonal pivotal cubic. See Bi-isogonal and Tri-isogonal Pivotal Cubics. Furthermore, this pK and its hessian generate a pencil of cubics which always contains a stelloid and a focal cubic which is the hessian of the stelloid. The radial center of the stelloid is the singular focus of the focal cubic. The pencil also contains two other K+ i.e. cubics with concurring asymptotes. With Ω = X(6), X(53), X(216), X(14576) we obtain K006, K049, K044, K415. The focal cubic associated to K006 is K1092. K627 is the barycentric product of H and K044 : for any point M on K044, the point H x M lies on K627. It is also the barycentric product of X(5) and K919. X(53) is a point of inflexion on the curve. The inflexional tangent is the line X(51), X(53) i.e. the line GK of the orthic triangle. The harmonic polar of X(53) is the common orthic axis of ABC and the orthic triangle hence K627 meets this orthic axis at three points with tangents concurring at X(53). These tangents are actually the symmedians of the orthic triangle.