   too complicated to be written here. Click on the link to download a text file.  X(3), X(182), X(3098), X(8666), X(8715) vertices of the CircumTangential triangle T1T2T3 vertices of the CircumNormal triangle N1N2N3 infinite points of K002    K735 is a central cubic with center O, meeting the circumcircle at the vertices of CircumTangential and CircumNormal triangles. It has three real asymptotes which are the parallels at O to those of the Thomson cubic K002. It meets K002 again at six (real or not) points which lie on the rectangular hyperbola (H) passing through X(3), X(54), X(140), X(1511), X(2574), X(2575). It meets the perpendicular bisectors again at six points which lie on the circle with center O, radius R/2, obviously inscribed in both equilateral triangles. K735 is the CircumTangential isogonal transform of K078, a stelloid with asymptotes parallel to those of K003. The CircumNormal isogonal transform of K735 is also a stelloid namely K736. See the similar cubics K1267, K1268, K1269, and Table 25. *** Generalization For any point P = u : v : w not lying at infinity, one can find a central cubic with center O, meeting the circumcircle at the vertices of CircumTangential and CircumNormal triangles and passing through the infinite points of pK(X6, P). Its equation is given     All these cubics form a net. Now, if P and Q are two points not lying on the circumcircle and not collinear with O then the central cubic K(P, Q) with center O, meeting the circumcircle at the vertices of CircumTangential and CircumNormal triangles, passing through P, Q and their reflections P', Q' about O meets the line at infinity at the same points as pK(X6, S) where S is the intersection of the lines L(P), L(Q) passing through the CircumTangential isogonal conjugates of P and P', Q and Q' respectively. The general equation is huge and will not be given here. K(P, Q) and pK(X6, S) meet again at six (real or not) finite points lying on a same rectangular hyperbola H(P, Q). Under CircumTangential and CircumNormal isogonal conjugations, K(P, Q) is transformed into two stelloids both with asymptotes parallel to those of the McCay cubic K003. Each stelloid is the reflection about O of the other. The radial centers XT and XN are the images of S under the homotheties with center O, ratios +1/3 and –1/3 respectively.   With P = X(182) and Q = X(8666), we obtain K735 and then P' = X(3098), Q' = X(8715), S = X(2), H(P, Q) = (H) as above. The two stelloids already mentioned above are K078 and K736. *** With P = X(3357) and Q = X(5450), we obtain K1267 and then P' = X(6759), Q' = X(6796), S = X(20). The two stelloids are the McCay cubic K003 and its reflection in O. *** In the figure opposite, P = X(6), Q = X(2), hence P' = X(1350), Q' = X(376). S is the intersection of the lines X(30)X(141) and X(511)X(8667) with SEARCH = -32.3886613230438. Note that the six points on H(P, Q) mentioned above are all real in this figure.     A special case Let P be a point on the Euler line and let pK(P) be the isogonal cubic with pivot P such that OP = T OH, a member of the Euler pencil, see Table 27. K(P) is the central cubic with center O, passing through the infinite points of pK(P), meeting the circumcircle at the vertices of CircumTangential and CircumNormal triangles. Under CircumTangential and CircumNormal isogonal conjugations, K(P) is transformed into two stelloids CT(P) and CN(P), both with asymptotes parallel to those of the McCay cubic K003. Each stelloid is the reflection about O of the other. K(P) and pK(P) meet again at six finite points O and Hi, i ∈ {1,2,3,4,5}, which lie on a rectangular hyperbola H(P) passing through X(3), X(54), X(2574), X(2575) and Q on the Euler line such that OQ = T / (1+T) OH. K(P) obviously contains the reflections of these five points about O. K(P) meets the perpendicular bisectors of ABC at three pairs of points {A1,A2}, {B1,B2}, {C1,C2}, which lie on a same circle C(P) with center O and radius R(T) = R √(T / (1+T)). These points are finite and real when T is not in ]-1,0[ i.e. when P is not between O and X(20). When P = O, K(P) is the union of the lines OTiNi , i ∈ {1,2,3}, and the six points coincide with O. H(P) is the Jerabek hyperbola. When P = X(20), three points are infinite and three coincide with O. pK(P) is K004 and K(P) is K1267. Note that C(P) and H(P) meet on the cubic K073.     When P traverses the Euler line, all these cubics form a pencil which is generated by two decomposed cubics : • the union of the altitudes of the CircumTangential and CircumNormal triangles, obtained with T = 0 hence P = X(3) and pK(P) is K003. • the union of the Euler line and the circumcircle, obtained with T = ∞ hence P = X(30) and pK(P) is K001. With T = 1/3 hence P = X(2), K(P) is K735 and pK(P) is K002. The following table gives a selection of cubics of this pencil (contributed by Peter Moses).  P K(P) X(i) on K(P) for i = pK(P) 2 K735 3, 182, 3098, 8666, 8715 K002 4 3, 1147, 7689 K006 5 3, 54, 7691 K005 20 K1267 3, 3357, 5450, 6759, 6796 K004 21 3, 1385, 3579 22 3, 15577, 44883 23 3, 12584, 32305 24 3, 9932, 9938 186 3, 12893, 12901 376 3, 4550, 8717 K243 550 3, 8718, 15062 1658 3, 16013, 45831 2071 3, 13289, 13293 5999 3, 511, 33388, 33389 K422 7488 3, 23358, 32401 7824 3, 8160, 8161 11286 3, 8667, 8716 14532 3, 8719, 9756 15329 3, 46608, 46616 22467 3, 22962, 22978   Two other remarkable examples • with P = X(15681), C(P) is the Stammler circle C(O, 2R). • with P = X(34350), C(P) is the 2nd Droz-Farny circle and H(P) passes through X(185), X(2904) .  