PROEESSOR CAYLEY ON THE BICIECULAR QUARTIC. 
455 
but £ 2 expressed as a function of v is an irrational function and would be 
the root of such a function ; so that if the form originally obtained had been this form 
$dv, it would have been necessary to transform it into the first-mentioned form 
(sfi+f) V€) ’ * n w ^ c k ^ expressed as a function of (x, y), that is of a. 
27. The system of course is 
= shdv + s 1 l 1 dv l + s 2 h 2 dv 2 + s 3 \dv 3 , 
dS l = zhdv — i l }) i dv l -j- e jH. 2 dv 2 -+- & 3 \dv 3 , 
dS 2 =zbdv — z$ l dv l — s f 2 dv 2 -j- s 3 \dv 3 , 
dS 3 = shdv — dv x — ejb 2 dv 2 — z 3 \dv 3 , 
where dv=— — ~ — r= &c. ; and this is the most convenient way of writing it. 
{x* + y*) V© J b 
Reference to Figure. — Art. No. 28. 
28. I constructed a bicircular quartic consisting of an exterior and interior oval with 
the following numerical data: 48, t /‘+$ 1 =56, 60, ^+^=80 ; 
g-\-6 3 = — 6, </+^, = 2, g-\-6 0 —Q, g-\- 4,= 26), — not very convenient ones, inasmuch as the 
exterior oval came out too large. The annexed figure shows 0, 1, 2, 3, the centres of the 
circles of inversion, the interior oval, and a portion of the exterior oval, also the origin 
and axes ; it will be seen that the centres 0, 2 lie inside the interior oval, the centres 
1, 3 outside the exterior oval: I add further the values 
y/f+0 3=6-93, \/— (#+0=2-45, a 3 =10T8, &=- -98, 
\/'f — - 7 - 48, \/g~\~Q\ =1*41, aj= 8*73, /3, = — |— 2*94, 
\//+ ^0=7*75, s/g-\-& o =2*45, a 0 = 8*15, /3 0 =+ '98, 
</AK= 8-94, Jg+S, =5-09, «,= 6-10, ft= + -23. 
We thus see how there exists a series of quadrilaterals ABCD, where A, B are situate 
on the interior oval, C, D on the exterior oval. Considering the sides as drawn in the 
senses A to B, B to C, C to D, D to A, and representing the inclinations measured from 
the positive infinity on the axis of x in the sense x to y , by y„ y 2 , y 3 , y respectively, then 
in passing to the consecutive quadrilateral A'B'C'IT, we have y, and v 2 decreasing, 
y 3 and y increasing, that is, dv x and dv 2 negative, dv 3 and dv positive ; then reckoning the 
elements AA, BB', CO, DD', that is dS„ dS 2 , dS 3 , dS, as each of them positive, we have 
dS 2 — ^S x =— 2& 1 dfy 1 , 
dS 3 — dS 2 = — 2\dv 2 , 
d$ — dS 3 = +2^ 3 <Zy 3 , 
<ZS, + dS =+2S dv, 
