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 W-isoconjugate 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 P1, P2, P3. 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.