1904.] BROOKS — ORTHIC CURVES. 309 



by selecting pairs in all possible ways from a, /5, y. 



The figure of nine points in which two orthic cubics intersect 

 may be regarded as an extension of the orthocentric four-point 

 determined by two equilateral hyperbolas. It is convenient to 

 extend the term orthocentric to such a figure. Resuming the results 

 obtained above, we have : 



U'he?i three of the points of an orthocentric nine-point are a triad 

 of the orthic curves through the nine points^ the remaining six points 

 are imaginary^ and are the antipoints of the three real poijits. The 

 centroid of the nine poi?its is the centre of every orthic cubic through 

 them. 



It is convenient to speak of a set of orthocentric points deter- 

 mined by a central pencil as a central set. Since any three points 

 determine a pencil of orthic cubics of which they are a triad, any 

 three points, with all their antipoints, form a central orthocentric 

 nine-point. 



XIV. The Foci. 



We shall now attack the problem of finding the foci of the orthic 

 cubic. Let us begin with a few words as to the way in which the 

 foci of a curve appear in analysis with conjugate coordinates. The 

 focus of a curve is the intersection of a tangent from one circular 

 point with a tangent from the other circular point. In other words, 

 if the circular rays from a point are tangent to a curve, that point 

 is a focus of the curve. Now the equation of the circular rays from 

 a point a, a, is' 



{x — a) {x — a) = o. 



Therefore, one of the'lines is , 



and the other is 



X — a =-= o. 

 Suppose the equation of the curve is 



F{xx) = o. 1 

 Now if the circular ray 



x — a = o 



