The collected mathematical papers of Arthur Cayley.

414]

ON POLYZOMAL CURVES.

535

similarly for finding the tangents at (r = 0, z =0) we have only to attend to the terms
of the second order in (7, z). 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 (q?, 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
/1 (- ( z - ( a') + \/m(3- z) ( -) + Vm ( - Z)) (v - y') = 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 (E, z); and it is easy to see that
any term involving a', /3', y' 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 /7, and throwing out this factor, it is
V1 (_- az) + Vmn ( -/ z) + Vn ( - yz) = 0,
viz., this equation which evidently belongs to a pair of lines passing through the point
(t=0, z=0) gives the tangents at the point in question; and similarly the tangents
at the point ( 7=0, z = 0) are given by the equation
/V (.t - a'z) V ( - /3'z) + V/ (r - - yz) = 0.
161. To complete the solution, attending to the tangents at ( =0, =0), and
putting for shortness
X = 1 -m -n,
/ = - + rn -n,
= -- - m +n,
A = 12 + m + n2 - 2mn - 2nl- 21m,
the rationalised equation is easily found to be
2.
- 2z (IXa + mr/3 + nvy)
+ z2 (12a2 + m2/2 + n272 - 2mnny - 2/nya - 2n 2ma) = 0;
and it is to be noticed that in the case of the circular cubic or when V+ V/m + Vn = 0,
then A =0, so that the equation contains the factor z, 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 (12a2 + m2/32 + n2y2 - 2mn/3y - 2nlyx - 21ma/3) - (1Xa + m/n,/ + nv7)2 = 0,
or observing that
A- X2 = - 4mn, &c.
A + iv =- 21X, &c.
this is
la2 + m/32 + nm2 + X/3 + wta + zva3 = 0,