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Let A and B be two circles, and AD, BF and AD, BE be parallel radii; if DF, DE be joined, they will cut AB in C and C", the internal and external centres of similitude.
For AC: BC :: AD : BF :: R : R', and AC" : BC' :: AD : BE :: R: R'.
Hence, C and C are the centres of similitude.
Corollary 1.-Any transversal through the centre of similitude is divided in the ratio of the radii.
Corollary 2.-Any tangent through the centre of similitude is divided in the ratio of the radii.
Corollary 3.--The line joining the centrès is divided harmonically at the centres of similitude.
Definitions.—The points E and D, as also M and N, are said to be homologous with respect to each other; and E and N, as also M and D, are said to be anti-homologous with respect to each other. If C'N, C'n be two secants, then Em, Nn are said to be anti-homologous chords.
Proposition 26. Theorem.—The product of the distances of a centre of similitude of two circles, from two anti-homologous points, is constant.
Let C" be a centre, of similitude of the circles A and B; then C'E. C'N is constant.
Let the power of C' with reference to the
circle A, or CD.C'N - P. Also (Prop. 25, Cor. 1), C'N R
Dividing the first by the second, C'D.C'M CM R'
R Now, D and M are anti-homologous points, and the second member of the equation is constant. Hence the theorem.
Corollary 1.—The anti-homologous chords meet on the radical axis.
Let Em, Nn be anti-homologous chords meeting at 0. Then, from the above,
hence (V. 27, Cor. 3), E, m, n and N are on the circumference of the same circle, and Em is a common chord of this new circle and the circle B; hence (Prop. 23, Cor. 1) it is a radical axis of these two circles. For the same reason Nn is a radical axis of the new circle and the circle A. Hence, O is the radical centre of the three circles, and OP (Prop. 24) the radical axis of A and B.
Corollary 2.-The extreme positions of the chords (that is, when they become tangents) also intersect on the radical axis.
Proposition 27. Theorem.-If three circles be given, and considered, taken two at a time, as forming three pairs
1. The external centres of similitude of the three pairs are in a straight line.
2. The external centre of similitude of any pair, and the internal centre of similitude of the other two pairs, are in a straight line.
Let A, B and C be three circles, C', C", C'', their external
centres of similitude, taken two and two; S and s', two internal centres of similitude; then will ("C" C" be a straight line, as also S, S', C'.
1. From (Sec. VIII., Def. 1), calling R, R', R" the radii of A, B and C respectively,
C'A R CC R" CB R
Multiplying these together,
But C'C" C'"' is a transversal cutting, the sides produced of the triangle ABC; hence (Prop. 9), it is a straight line.
Hence, SA.S'C.C'B=SC.S'B.C'A, and (Prop. 9) S, S' and Care in a straight line.
In the same way, S", S, C"" is a straight line, and also S'S'C".
Definitions.-The line CC"" is called the external axis of similitude; and C", S', S is called the internal axis of similitude.
Proposition 28. Theorem.--If a circle touch two other circles, the chord of contact will pass through a centre of similitude.
Let the circle A touch the circles B and C in a and b. The chord ab will pass through S, a centre of similitude of B and C.
The lines AC, BA will pass through the points of contact, b and a. Produce ab to c; join Cc, Bd. Then, because the isosceles triangles Bda, Ccb, Aba are similar, the radii Ba, Cc are parallel. Therefore the chord ab passes through (Prop. 25) a centre of similitude S.
Proposition 29. Problem. —To describe a circle tangent to three given circles.
There may be eight circles tangent to the three circlesone touching all three externally, one all three internally; and three other pairs, one of each having internal contact with one circle, and external with two; and the other having internal contact with two, and external with one.
Let A, B, C be the three given circles, c"C" C"" the external axis of similitude.
Suppose the problem solved, and let ad', 66, cd be the three chords of contact of the circles M and N.
We will prove
1. That these chords pass through the radical centre of the three given circles.
. For, a being an internal centre of similitude (Sec. VIII., Def. I., Cor.) of the circles A and M, and a' of A and N, the chord aa' will pass through (Prop. 28) the external centre of similitude of the circles M and N. For the same reasons will also bb' and cc'. Hence, they meet in R. Also (Prop. 28), ab, a'b' will pass through the external centre of similitude C' of the circles A and B. Hence, aa', bb' are anti-homologous chords (Prop. 25, Def.), and their intersection R is (Prop. 26,