414] 
ON POLYZOMAL CURVES. 
535 
similarly for finding the tangents at (77 = 0, 2 = 0) we have only to attend to the terms 
of the second order in (77, 2). But it is easy to see that 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 (77, 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 
VZ (£ — az) (77 — a'z) + \/m (£ — /3z) (77 — /3'z) + Vn (£ — 72) (v — y'z) = 0. 
For finding the tangents at (£ = 0, 2 = 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', 7' will be of the third order at least in (£, z), that is, 
we may reduce a, ¡3', y each to zero; the irrational equation then becomes divisible 
by V77, and throwing out this factor, it is 
\!l (£ — az) + Vm (£ — (3z) + Vn (£ — 7z) = 0, 
viz., this equation which evidently belongs to a pair of lines passing through the point 
(£ = 0, 2 = 0) gives the tangents at the point in question; and similarly the tangents 
at the point (77 = 0, 2 = 0) are given by the equation 
Vl (77 — a'z) + Vm (77 - f3'z) + Vn (77 — y'z) = 0. 
161. To complete the solution, attending to the tangents at (£=0, 2=0), and 
putting for shortness 
X = l — m — n, 
¡x = — l + m —11, 
v = — l — m + n, 
A = l 2 + vr& + v? — 2 mn — 2nl — 2lm, 
the rationalised equation is easily found to be 
r-A 
— 2 £2 (IXa + vifx/3 + nvy) 
+ 2 2 (Pcl 1 + m 2 /3' 2 + n 2 7 2 — 2mn(3y — 2nlya — 2lma(3) = 0 ; 
and it is to be noticed that in the case of the circular cubic or when vT+ Vm + V«= 0, 
then A = 0, so that the equation contains the factor 2, and throwing this out, the 
equation gives a single line, which is in fact the tangent of the circular cubic. 
162. Returning to the bicircular quartic, we may seek for the condition in order 
that the node may be a cusp: the required condition is obviously 
A (l 2 a 2 + mrft 2 + n 2 7 2 — 2mn/3y — 2nlyx — 2lmaf3) — (l\a + m/x/3 + nvy) 2 = 0, 
or observing that 
this is 
A — X 2 = — 4tmn, &c. 
A + /jlv = - 21\, &lc. 
la? + mfi 2 + ny 2 + X/37 + /t ya + va(3 = 0,