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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 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 in-conic with center K. Thus, all these cubics form a pencil of cubics. The general barycentric equation is : |
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with singular focus F = a^2(q-r) : : , point at infinity Inf = (q-r) : : , real asymptote 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. Further details in CL001.
Table 4 shows a selection of nK0(X6,R). |
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