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X(15), X(16), X(110), X(526) centers of the 3 Apollonius circles 

Given an isoconjugation with pole W(p : q : r), the locus of point P such that the pedal triangles of P and its Wisoconjugate P* are parallelogic is nK0(W, X6). This cubic is circular if and only if W = X(50). K148 is a member of the class CL022 of cubics. It singular focus is X(23) and its real asymptote is the line X(323)X(526). K148 is the locus of M such that the Euler line of the pedal triangle of M is perpendicular to the Euler line. See the related K292, K640, K641. See also Orthopivotal Cubics. K148 is the isogonal transform of K064. 

K148 has always three real prehessians denoted P_{1}, P_{2}, P_{3}. The centers of the polar conics of X(526) with respect to these prehessians are X(110), X(15), X(16). Furthermore the polar conic of X(526) in K148 is a rectangular hyperbola. It follows that K148 is the isogonal pK with pivot X(526) with respect to the triangle X(15)X(16)X(110). Thus it must pass through the in/excenters of this latter triangle with tangents parallel to the real asymptote. *** Additional remarks : • the orthic line of K148 is the tangent at X(110) to the circumcircle. Hence, the polar conic of any of its points is a rectangular hyperbola. • K148 meets its real asymptote at X = X(9213), the antipode of X(23) on the Parry circle. X is the isogonal conjugate of X(526) in the triangle X(15)X(16)X(110). Hence, the tangents at X(15), X(16), X(110) pass through X. • the X(50)–isoconjugate of X is the intersection X* = X(5467) of the Brocard axis and the orthic line. *** More generally, if M is a point on K148 which is not a flex and if M1, M2, M3 are the centers of the polar conics of M with respect to the prehessians then K148 is a pK with respect to the triangle M1M2M3. Its pivot is M and its isopivot is the tangential M' of M. For example, when M = X*, we obtain M1 = X and the polar conic of M is a rectangular hyperbola. 
