414] 
ON POLYZOMAL CUR YES. 
571 
tangent circles; viz., given any three circles A, B, C in the same plane, to draw the 
tangent circles. Taking the antipoints of the three circles, then selecting any three 
antipoints (one for each circle) so as to form a triad, we have in all four complementary 
pairs of triads. Through a triad, and through the complementary triad draw two 
circles, these are situate symmetrically on opposite sides of the plane; and combining 
each antipoint of the first circle with the symmetrically situated antipoint of the second 
circle, we have two pairs of points, the points of each pair being symmetrically situate 
in regard to the plane, and having therefore an anticircle in this plane; these two 
anticircles are a pair of tangent circles; and the four pairs of complementary triads 
give in this manner the four pairs of tangent circles. 
I return to the equations 
(cd - S) 2 + {y - S') 2 + (z - S") 2 = 0, 
(a - S) 2 + {a' - SJ + {a" - 8" ) 2 = 0, 
(b - 8y 4- {V - S'y + (b" - S") 2 - 0, 
(c - Sy + (c' - S'y + (c" - S") 2 = 0 ; 
by eliminating (S, S', S") from these equations we shall obtain the equation of 
the pair of cone-spheres through the points (a, a', a"), (b, b', b"), (c, c', c"). Write 
cd — S, y — S', z — S" = X, Y, Z, then we have X 2 + F 2 + Z 2 = 0, and, putting for shortness 
SI = (a — x) 2 + {a' — y) 2 + (a" — z) 2 , 
33 = (b - x) 2 + (b' - y) 2 + (b" - z) 2 , 
(£ = (c — x) 2 + (c' — y) 2 + (c" — z) 2 . 
then, by means of the equation just obtained, the other three equations become 
SI + 2 [(a — x) X + (a' — y) F+ (a" — z) Z] = 0, 
S3 + 2 [(b-x)X + (b' — y)Y+ (6" -z)Z] = 0, 
g +2[(c-x)X + (c'-y) Y+(c"-z)Z] = 0. 
These last equations give 
Z : F : ASl + 8 +v g 
: VSl +/i'S3 + */g 
: V'Sl + //'S3 + v'% 
where 
A = b'c" — b”c' + (c' —b')z— (c" — b") y, 
H — c'a" — c"a! + (a' —c' )z — (a" — c")y, 
v = a'b" - a"b' + (6' — a!)z — (b" - a") y, 
X' = b"c -be" +(c" -b")x-(c -b )z, 
y! = c'a — ca" + {a" — c")x — (a—c )z, 
v = a'b — ab" + (b" — a") x — {b —a ) z, 
be' —b’c + (c — b )y — (c —b' )x, 
y" = ca' —c'a + (a —c )y — (a' — c' )x, 
v" = ab' —a'b+(b —a)y — (b—a')x; 
72—2