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
