In the same way, S", S, C"" is a straight line, and also S"S' C".

Definitions.-The line C'C"C"" 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, Ce 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 aa', bb', ce' 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,

Cor. 1) a point in the radical axis of A and B. For the same reasons it is in the radical axis of B and C, and of A and C, and is therefore the radical centre of the three circles.

2. Each of the chords aa', bb', cc' contains the pole, with reference to its own circle, of the external axis of similitude C'C"C""

For, since R is in the chord of contact of the circle A with the circles M and N, it is (Prop. 28) a centre of similitude of M and N. Hence the two chords ab, a'b' are anti-homologous, and their point of meeting, C', is (Prop. 26, Cor. 1) a point of the radical axis of the circles M and N.

Similarly, it could be shown that this radical axis passes through C" and C'""', and therefore agrees with the external axis of similitude. Hence the line C'C" C'""' contains (Prop. 26, Cor. 2) D, the point of meeting of the two anti-homologous tangents Db, Db'; that is, D is the pole (Prop. 20, Cor. 3) of the chord bb' with reference to the circle B; hence (Prop. 21) the chord bb' contains the pole of the polar C" C" C'"' with reference to the circle B. Also aa', ce' contain the poles of C"C" C"" with reference to A and C.

Consequently, we have the following rule to solve our problem:

Determine the radical centre, R, and the external axis of similitude of the three circles, A, B, C; take, with reference to each of them, the poles P, P', P" of this axis, and draw the lines RP, RP, RP", cutting the circles respectively in a and a', b and b', c and c'; through the points a, b, c pass a circle; it will be the circle of internal contact; through a'b'c' pass a circle; it will be the circle of external contact.

By substituting for the external axis of similitude the internal axes, the other tangent circles may be found in a similar way.