46 
NOTE ON THE CARTESIAN. 
[565 
the other two points P a , P c . I propose to find the two Cartesians in question. To 
fix the ideas, let the points G, B, A be situate in order as shown in the figure, their 
distances from a fixed point 0 being a, b, c, so that writing a, /3, y = b — c, c — a, a — b 
respectively, we have a + /3 4- y = 0, and a, y will represent the positive distances GB 
and BA respectively, and — ¡3 the positive distance AG. Suppose, moreover, that the 
distances PA, PB, PC regarded as positive are R, S, T respectively; and that the 
distances P b A, P b B, P b G regarded as positive are R', S', T' respectively. 
Suppose that for a current point Q the distances QA, QB, QG regarded as 
indifferently positive, or negative, are r, s, t respectively; then the equation of a 
bicircular quartic having the points A, B, G for axial foci is 
Ir + ms + nt = 0, 
where l, m, n are constants; and this will be a Cartesian if only 
P m 2 n 2 n 
-+-0 + - = 0. 
a ß y 
We have the same curve whatever be the signs of l, m, n, and hence making the 
curve pass through P, we may, without loss of generality, write 
IR + mS + nT = 0, 
R, S, T denoting the positive distances PA, PB, PC as above. We have thus for 
the ratios l : m : n, two equations, one simple, the other quadric; and there are thus 
two systems of values, that is, two Cartesians with the foci A, B, G, and passing 
through P. 
I proceed to show that for one of these we have — IR' + mS' 4- nT' = 0, and for 
the other IR’ + mS' — nT' = 0, or, what is the same thing, that the values of l : m : n are 
l : m : n = — (ST' + S'T) : TE + T'R : RS' - R'S, 
and 
l : m : n— (ST - S'T) : - (TE + T'R) : RS' + R'S; 
viz. that the equations of the two Cartesians are 
r , 
s , 
t 
= 0, and 
r , 
s , 
t 
R, 
s, 
T 
R, 
s, 
T 
-E, 
S', 
T 
E, 
S', 
-r