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Given a point R = u:v:w (the root of the cubic), the isogonal nK0(X6, R) has barycentric equation : ∑ u x (c^2 y^2 + b^2 z^2) = 0. It always passes through the intercepts U, V, W of the trilinear polar of R on the sidelines of ABC. The points A and U, B and V, C and W share the same tangentials A', B', C' respectively which are three collinear points on the cubic. Any nK0(X6, R) can be considered as the locus of M such that M and its isogonal conjugate M* are conjugated with respect to :
Note that the cubic contains the vertices of the diagonal triangle of the four basis points of the pencil although these are not necessarily all real. See a figure below and also K082. *** Circular nK0(X6, R) nK0(X6, R) is circular if and only if R lies on the orthic axis. It is always a focal cubic with singular focus F on the circumcircle, isogonal conjugate of the point at infinity of the trilinear polar L of the isotomic conjugate of R. This line L passes through X(69) = isotomic conjugate of the orthocenter H. Such a cubic is the locus of foci of inscribed conics centered on a line passing through K called the axis or the orthic line of the cubic. This axis passes through the midpoints of AU, BV, CW. Hence the cubic always contains A, B, C, the circular points at infinity and the (four) foci of the inconic with center K. Thus, all these cubics form a pencil of cubics. The general barycentric equation is : ∑ (a^2 p + SB q + SC r) x (c^2 y^2 + b^2 z^2) = 0 where p:q:r is any point except the centroid G. Its singular focus is F = a^2(qr) : : , its point at infinity is Inf = (qr) : : , its real asymptote is the homothetic of the axis under h(F,2). See CL025.
nK0+(X6, R) nK0(X6, R) has concurring asymptotes if and only if R lies on the cubic K138. This cubic contains X(2) giving K082 (Equal power+ cubic) and X(6) giving Kjp = K024.
Equilateral nK0(X6, R) nK0(X6, R) is equilateral if and only if R is the Lemoine point K = X(6). This is Kjp = K024.
Nodal nK0(X6, R) nK0(X6, R) is a nodal cubic if and only if it contains one and only one of the in/excenters which is therefore the node. If we take the incenter I = X(1), R must lie on the antiorthic axis (trilinear polar of I). The nodal cubics with node I form a pencil of cubics which contains only one focal cubic when R = X(650) (Pelletier point). It is the Pelletier strophoid K040.
Central circular nK0(X6, R) nK0(X6, R) is a central circular cubic if and only if its center (and singular focus) is one of the three (always real) points where the Grebe cubic K102 meets the circumcircle. These are the vertices of the Grebe triangle. Further details in CL001. 



The figure shows nK0(X6, X9) with some conics defined above. It is the locus of M such that M and its isogonal conjugate M* are conjugated with respect to any of the conics of the pencil generated by (C) and (H). This cubic meets the circumcircle again at A', B', C' whose Simson lines concur at X, the midpoint of HΩ, where Ω is the center of the circle of the pencil. Indeed, the cubic is also the locus of M whose pedal circle is orthogonal to the circle with diameter HΩ. 
