Full text: The collected mathematical papers of Arthur Cayley, Sc.D., F.R.S., sadlerian professor of pure mathematics in the University of Cambridge (Vol. 6)

414] 
ON POLYZOMAL CURVES. 
551 
189. These values of V7 : Vm : Vn give the equations of the two circular cubics 
with the foci (A, B, G, D), the equation of each of them under a fourfold form, 
viz., we have 
and 
( 
• 
) 
d 1 
^1 > 
61- 
di, 
Ci - 
di, 
• } 
d\ 
(h) 
d x - 
CL] 
d 1} 
• 
) 
&i- 
Ci, 
Cl 
-a 1, 
eli 
■b 1} 
> 
- 
Cl 
— d l , 
di 
+ bji 
dl 
+ Ci, 
. 
5 
Cli 
di 
b 2 
— di, 
d 1 
cii> 
bl 
Ci 
Ci 
+ CLi, 
— Ui 
-61 
cii - c x 
— Uj 
(first curve), 
&i + c x )(V2l, VS, Ve, V'D) = 0 
Ci — cq 
(second curve). 
190. Similarly GA and BD meet in S, and if we denote by a 2 , b. 2 , c. 2 , d 2 the 
distances from S of the four points respectively, so that c 2 a 2 = b 2 d 2 = rad. 2 S (observe 
that if as usual A, B, G, D are taken in order on the circle 0, then A, G are on 
opposite sides of S, and similarly B, D are on opposite sides of S, so that taking 
«2j b. 2 positive c 2 , d 2 will be negative), we have 
a : b : c : d = c 2 (b 2 — d 2 ) : d 2 (c 2 — a 2 ) : — a 2 (b 2 — d 2 ) : — b 2 (c 2 — a 2 ), 
and then the equations ^- + ^ + - = 0, Vi + Vm + V?i = 0, are satisfied by the two sets 
of values 
Vi : Vm : Vn= b 2 — c 2 : c 2 — a 2 : a 2 — b 2 , 
and 
= — b 2 — c 2 : c 2 — a 2 : a 2 + b 2 , 
and we have the equations of the same two cubic curves, each equation under a 
fourfold form, viz., these are 
and 
191. 
distances 
( . , 
- c 2 + do, 
— d 2 J rb 2 , — 6 2 + c 2 
(V21, VS, VS, Vî))=0 
c 2 — do, 
• > 
d 2 —a 2 , — c 2 + a 2 
— b 2 + d 2 , 
tt 2 — do, 
, — a 2 + 60 
(first curve), 
b 2 — c 2 , 
Cq CL 2) 
a 2 — i> 2 , 
( • ’ 
c 2 + do, 
— d 2 + 6 2 , —b 2 — c 2 
) (Vi, vs, ve, V®) = 0 
d 2 Co, 
• ) 
a 2 + d 2 , c 2 —a 2 
— 60 + d 2 , 
— d 2 — ci 2) 
, a 2 +b 2 
(second curve). 
bo + c 2 , 
— C 2 + Cl 2 , 
— a 2 — b 2 , 
And again 
Ai? and 
&D meet in T, and denoting by a 3 , 6 3 , c 3 , d 3 
from T of the four points respectively, so that a 3 b 3 = c 3 d 3 = rad. 2 T, we have 
a : b : c : d = b 3 (c 3 — d 3 ) : — (i 3 (c 3 — d 3 ) : d 3 (u 3 b 3 ) : c 3 (u 3 6 3 ) 5 
the
	        
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