MB. R. LACHLAN ON SYSTEMS OF CIRCLES AND SPHERES. 
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and, therefore, the discriminant is 
( 'Jc-\-a)(Jc-{- b) { (/c+c)(/j+ d) — u 3 } = 0, 
which can only have equal roots when 
(c-\-d) 3 — 4c<i+#=0. 
If n be real this can only be satisfied, by n— 0, c—d. 
But of our four circles of reference one must be imaginary; if one of the circles 
[x, y) is imaginary n will be real, and our equation will reduce to 
ax 2 -\-by 3 -\- cz 3 + civ 3 =0, 
which represents a pair of circles, and need not be considered. 
If both circles ( x , y) are real, then it is simplest to take as system of reference, these 
circles and their two points of intersection. So that the absolute is of the form 
x 3 -\-y 3 =izw, 
and the discriminant becomes 
( Jc-\-a)(k-{-b) \cd—(2Jc—n) 3 } = 0 ; 
and, then, if this has equal roots, either c = 0 or d— 0 ; and the equation takes the 
form 
ax 3 -f- by 2 + cz 3 +2 n zw= 0, 
which, by means of the absolute, may be written 
a x 3 + by 3 + cz 3 — 0. 
Let us suppose now that three of the roots of the discriminant are equal; referring 
our coordinates to the circle corresponding to the unequal root, and any three circles 
cutting it and one another orthogonally, the equation of the curve will reduce to 
ax 3 -\-by 3j rCz 3J (-dw 3 -\- 2 fyz + 2mytv + 2 nzw = 0 ; 
and the discriminating cubic is 
(k-\-a) &+&, f m =0, 
f, k+c, n 
m, n, h-\-d 
which we can easily prove can only have three equal roots when 
f=m=n= 0, b=c=d ; 
provided that f m, n are all real, in which case the circle x=0 is imaginary. 
In this case the curve takes the form 
which represents a point. 
ax 3 -\-by 3 -j- bz 3 ~\~bw 3 = 0, 
