PROFESSOR CAYLEY ON POLYZOMAL CURVES. 87 



II. We may have 



there is in this case a single branch ideally containing (z = 0) the line infinity ; 

 the order is = 7. Each of the points I, J is a triple point, there is consequently 

 one other point at infinity; viz., this is a real point, or the curve has a real 

 asymptote. There are 6 nodes as before ; dps. are 6 + 2 . 3, = 12 ; class = 18, 

 deficiency = 3. 



III. We may have 



J I + Jm — 0, Jn + Jp = ; 



there are then two branches each ideally containing (z = 0) the line infinity; the 

 order is = 6. Each of the points /, J is a double point, and there are therefore 

 two more points at infinity. These may be real or imaginary ; viz., the curve 

 may have (besides the asymptotes at /, J) two real or imaginary asymptotes. 

 The circles Jlfk + JmB = 0, JnG + JpD — 0, each contain (z = 0) the line 

 infinity, or they reduce themselves to two lines, so that in place of two nodes we 

 have a single node at the intersection of these lines ; number of nodes is = 5. 

 Hence dps. are 5 + 2.1,= 7. Class is = 16, deficiency = 3. 



IV. We may have 



J I : Jm '■ Jn : Jp = a : b : c : d 



there is here a single branch containing (z 2 = 0) the line infinity twice ; the 

 order is = 6. Each of the points /, J is a double point, and there are therefore 

 two more points at infinity, that is (besides the asymptotes at i, J), there are 

 two (real or imaginary) asymptotes. The number of nodes, as in the general 

 case, is = 6. Hence dps. are 6 + 2 . 1, = 8 ; class is = 14; deficiency = 2. 



I notice the included particular case where the circles reduce themselves to 

 their centres; viz., we have here the curve 



a JA~+ b JW+ c ^0"+ d JD~= , 



which (see ante No. 93) is in fact the curve which is the locus of the foci of the 

 conies which pass through the four points A,B, C,D. It is at present assumed that 

 the four points are not a circle; this case will be considered post No. 199. 

 If we have BC, AD meeting in R ; CA, BD in S, and AB, CD in T, then these 

 points R, S, T are three of the six nodes. In fact, writing down the equations 

 of the two circles 



hjB + Cn /C = 0, &JK+ djD~= 0, 



and observing that when the current point is taken at R, we have B : C 

 = KB 2 : RC 2 = (BAD) 2 : {CAD) 2 = c 2 : b 2 , and similarly A : D = RA* : RD 1 = 

 {ABC) 2 : {DBC) 2 = d 2 : a 2 , we see that each of the two circles passes through 



