PROFESSOR CAYLEY ON POLYZOMAL CURVES. 67 



By what precedes, the curve is of the order = 4, touching each of the given 

 circles twice, and having a double point, or node, at each of the points /, J; that 

 is, it is a bicircular quartic : but if for any determinate values of the radicals 

 s/l, s/m, s/n, we have 



then there is a branch 



JW + y/mB° + J^° = 0, 



containing (z — 0) the line infinity ; and the order is here = 3 : viz., the curve 

 here passes through each of the points /, J and through another point at infinity 

 (that is, there is an asymptote), and is thus a circular cubic. 



160. I commence by investigating the equations of the nodal tangents at 

 the points /, J respectively ; using for this purpose the circular co-ordinates 

 (£, tj, s = 1), it is to be observed that, in the rationalised equation, for finding the 

 tangents at (£ = 0, z = 0) we have only to attend to the terms of the second 

 order in (£, s), and similarly for finding the tangents at (>? = 0, z = 0) we have 

 only to attend to the terms of the second order in (>?, z). But it is easy to see 

 that on any term involving a", b", or c" will be of the third order at least in (£, z), 

 and similarly of the third order at least in (>?, z) ; hence for finding the tangents 

 we may reject the terms in question, or, what is the same thing, we may write 

 a", b", c" each = 0, thus reducing the three circles to their respective centres. 

 The equation thus becomes 



Jl(% - az) (ti - a'z) + Jm(% - jfe) (, - #») + Jn(& - yz) - y'z) = 0. 



For finding the tangents at (£ = 0, z = 0) we have in the rationalised equation to 

 attend only to the terms of the second order in (£, z) ; and it is easy to see that 

 any term involving a, (3 ', </ will be of the third order at least in (£, z), that is, 

 we may reduce a', j3', y each to zero ; the irrational equation then becomes 

 divisible by */>/, and throwing out this factor, it is 



J I {I - *z) + Jm{l - /&) + Jn(& - yz) = , 



viz., this equation which evidently belongs to a pair of lines through the point 

 (£ = 0, z = 0) gives the tangents at the point in question ; and similarly the 

 tangents at the point (>/ = 0, z = 0) are given by the equation 



s/l(y — a'z) + As/ ,?i (i — @' z ) + s/n{n — yz) = 0. 



161. To complete the solution, attending to the tangents at (£ = 0, z = 0), 

 and putting for shortness 



X = I — m — n , 



/j, = — I + m — n , 



v = — I — m + n , 



A = I 2 + m 2 + v? — 2mn — 2nl — 2lm, 



