Pedal and Antipedal Circles tangent

Are there real points P such that the pedal and antipedal circles of P are tangent?

Antreas P. Hatzipolakis, Hyacinthos #21746

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I get the cubic

S^2 xyz + CyclicSum[ a^2 y z (c^2 y + b^2 z)] = 0.

Francisco Javier, Hyacinthos #21747

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Yes, if P is on an octic or on the cubic K191="circumcircle pedal cubic, nK(X6, X6,?)".

If P is on the cubic K191, then the point of the contact of the two circles are on the circumcircle.

K191: S^2 xyz + CyclicSum[ a^2 y z (c^2 y + b^2 z)] = 0,

or equivalently

K191: S^2 xyz + CyclicSum[ a^2 x (c^2 y^2 + b^2 z^2)] = 0.

S= 2*area(ABC) (In CTC of Bernad Gibert, S=area(ABC))

Angel Montesdeoca, Hyacinthos #21750

The cubic S^2 xyz + CyclicSum[ a^2 y z (c^2 y + b^2 z)] = 0 is not K191

It will be K634 in Bernard Gibert's list