both cubics : X(2), X(4)
K070a (outer cubic) = pK(X4, X1586) : X(2), X(4), X(486), X(492), X(1586), X(3069), X(13390)
K070b (inner cubic) = pK(X4, X1585) : X(2), X(4), X(485), X(491), X(1585), X(1659), X(3068)
Let P be a point and PaPbPc be its cevian triangle. Erect outwardly or inwardly an arbelos (shoemaker's knife) on each side of triangle ABC and inscribe a circle centered at Oa having Xa as contact with the semi-circle with diameter BC. See figure 1 and also Peter Woo, Simple Constructions of the Incircle of an Arbelos, Forum Geometricorum, vol.1 (2001), pp. 133 -136.
Define similarly Ob and Oc, Xb and Xc.
The locus of point P such that ABC and XaXbXc are in perspective is a shoemaker's cubic.
They are anharmonically equivalent to K006.