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X(5000) and X(5001) are the antiorthocorrespondents of the Lemoine point X(6). These points are real if and only if ABC is acutangle. Their midpoint is X(468) and their barycentric product is X(232). Their perpendicular bisector is the orthic axis. They lie • on the Euler line, • on the circumconic with perspector X(3569) passing through X6, X232, X250, X262, X264, X325, X511, X523, X842, X1485, X2065, • on the diagonal conic passing through X6, X157, X230, X523, X1503. • on the circles which are the antiorthocorrespondents of the lines passing through X(6). These circles are centrered on the orthic axis and are orthogonal to the circumcircle and more generally to all the circles mentioned below. Note that a line meets the associated circle on K018 since K018 is the orthopivotal cubic O(X6). They are inverse in any circle of the coaxal system containing the circumcircle, the ninepoint circle, the orthocentroidal circle, the orthoptic circle of the Steiner inellipse, the polar circle (hence they are orthoassociate points), the tangential circle. Recall that the radical axis of this system is the orthic axis. The following table shows a selection of listed curves passing through these two points (which are not repeated in the table). Some of these curves are generalized below. 



Note 1 : more generally, any nK0(Ω, X523) with pole Ω on the line X6, X25 contains X5000, X5001 and also X6. In particular nK0(X2393, X523) is a K+. K018 is the only circular cubic of this pencil. Note 2 : more generally, any pK(Ω, X98) with pole Ω on the line X4, X32 contains X5000, X5001 and obviously X98. K336 is the only circular cubic of this pencil. Note 3 : more generally, any pK(X232, P) with pivot P on the Euler line contains X5000, X5001 and the four square roots of X232. Recall that X232 is the barycentric product of these two points. K337 is the only circular cubic of this pencil. Note 4 : more generally, any nK0(Ω, X6) with pole Ω on the orthic axis contains X5000, X5001 and also X523. K608 is the only K+ (and even K++) of this pencil. Note 5 : more generally, a circular psK passing through X5000, X5001 must have • its pseudopole on K782 = pK(X2211, X25) through X(i) for i = 4, 6, 25, 232, 237, 248, 511, 694, 3053, 3186, 3563, • its pseudopivot on K780 = pK(X4, X297) through X(i) for i = 2, 4, 69, 98, 230, 297, 393, 694, 1503, 1987, 6330, 9473, • its pseudoisopivot on K783 = pK(X237, X3) through X(i) for i = 2, 3, 6, 154, 232, 237, 511, 1297, 1976, 1987, 3164, 9292, 9475. Remark that some of these psKs are in fact pKs. Note 6 : the following nK0s also contain X5000, X5001 : • any nK0(X232, P) with root P on the line X6, X523, • any nK0(X325 x X2501, P) with root P on the line X6, X264, • any nK0(X237 x X648, P) with root P on the line X250, X523. Note 7 : more generally, a circular nK passing through X5000, X5001 must have • its pole on nK0(X2211, X25) through X(i) for i = 6, 112, 232, • its root on nK0(X4, X297) through X(i) for i = 2, 4, 523, 648. Remark that some of these nKs are in fact nK0s. Note 8 : all these circular quartics are in a same pencil which also contains the union of K018 with the line at infinity. Note 9 : Q037 and Q054 are both bicircular quintics which generate a pencil of quintics of the same type. One is decomposed into K018 and the line at infinity counted twice. Note 10 : Q021 and Q024 are both bicircular sextics which generate a pencil of sextics of the same type. One is decomposed into Q019 and the circumcircle. 
