Full text: The collected mathematical papers of Arthur Cayley, Sc.D., F.R.S., late sadlerian professor of pure mathematics in the University of Cambridge (Vol. 13)

921] 
ON SOME PROBLEMS OF ORTHOMORPHOSIS. 
201 
of figure 5 : and similarly to DA, AB and BG of figure 4 the remaining quarter- 
circumferences DA, AB, BG of figure 5; that is, to the whole boundary of the square 
in figure 4 there corresponds the whole circumference of the circle in figure 5. 
14. To any line in figure 4, parallel to the axis Ox or to the axis Oy, there 
corresponds in figure 5 a bicircular quartic of the form x? + y 2 - ZAx-f - 2By? —1=0. 
The investigation is substantially the same as that contained in No. 12, and need 
not be here given. But it is remarkable that also, to any line of figure 4 parallel 
to a side of the square (that is, to any line x + y = c), there corresponds in figure 5 
a bicircular quartic of the like form (for the sides of the square, or lines x ±y = ±\ix 
of figure 4, this bicircular quartic becomes the twice repeated circle x? + y* - 1 = 0 
of figure 5, which is the result just obtained in No. 13). I investigate this result 
as follows. Writing, as in No. 13, 
snl x=p, snl iy = i snl y = %q, 
we have, as above, 
p 
x,= - 
Vi - 
q V1 — p A 
p 2 + q 2 
t 2/i = i » and thence x 1 2 + y? = -p— 
1 — p 2 q 2 1 — p 2 q 2 1 — 
p 2 q 2 ' 
c = 
Xi+y 1 
Now assuming between x and y the relation x + y = C, and writing snlO=c, this gives 
_p ^ 1 — q 4 + q Vl —p 4 
1 + p 2 q 2 
to obtain the required curve, we must between these equations (three independent 
equations) eliminate p and q. We have 
_ p Vl — q 4 + q Vl — p* 
1 — p-q 2 
and consequently 
(1 + p 2 q 2 ) c = (1 -py) {x 1 + y x ), 
or, writing for convenience 0=1 — p 2 q 2 , this equation gives 
0 = — . 
+ Vl + C 
Hence £lx 1 = p Vl — q 4 , Dy 1 = q\/l -p 4 ; these equations may be written 
Gl 2 x? = p 2 — (1 — O) q 2 , 
£l 2 y? = — (1 - O) p 2 + q 2 , 
and from these equations obtaining the expressions for p 2 and q 2 , and thence the 
expression for p 2 q 2 , =1—0, we find, after some easy reductions, 
But we have 
{Oi 2 + Vif - 1} + x iVi = —^2 ■ — ■ ■ 
O 2 
4c 2 
1-0 (®! + y^f - c 2 ’ 
or substituting this value, the equation becomes 
. , , 4c 2 «i 2 y 1 2 («1 + yi) 2 -i 
^ 1 + Vl ’ + {x 1 + y?f - c 2 C 2 
c. XIII. 
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