determined by the equation AB.r,cos y = ±BC.r±CA.r2; like signs being taken when the contacts are similar and unlike signs when the contacts are dissimilar. The four possible values arising from the selections of sign on the right side of the equation give the values of y corresponding to each assigned species of contact.

COR. 3. In Cor. 2, if cos y=0, the centres C of the particular circles of the system which are cut at right angles are given by the relation

or

BC.r1±CA.r2=0,
AC/BC= ±r1/r2

Hence, the variable circle having similar contacts with two given circles cuts at right angles the coaxal circle whose centre is their external centre of similitude; and, if the contacts are dissimilar, the coaxal circle whose centre is the internal centre of similitude.

COR. 4. If a= ±ẞ and y=90, the equation reduces to AC/BC= ±r1/r2, as in Cor. 3. Hence, the variable circle cutting two others at equal or supplemental angles cuts at right angles their external or internal circle of antisimilitude respectively.

COR. 5. Let the radius of the variable circle be infinite; hence (Cor. 3) all lines cutting two circles at equal or supplemental angles are diameters of their external or internal circles of antisimilitude.

EXAMPLES.

1. To describe a circle cutting any three circles A, r1; B, r2 ; C, r at given angles a, ẞ, y.

[The required circle cutting B, r1⁄2 ; C, r ̧ at given angles, therefore touches a known circle coaxal with them by Cor. 2; similarly for each of the remaining pairs of the given circles; hence the problem

reduces to "describe a circle touching three given circles with assigned contacts." There are in consequence eight solutions. These are given in a subsequent chapter.]

2. Show that Ex. 1 cannot be reduced to describing a circle cutting three given circles orthogonally.

[For let X be the circle coaxal with B and C which is cut orthogonally by the required circle, and constructed by putting y=90 in the relation of the present Article; similarly let Y coaxal with C and A, and Z coaxal with A and B, be circles cut orthogonally by it. Their centres, being found by the relations

are collinear, Art. 62, and their common orthogonal circle therefore indeterminate.]

3. A variable circle P, p touches two others A, r1; B, r2; show that the square of the common tangent t, to it and any third circle C, r3 coaxal with them, varies as its radius (t2 ∞ p).

་

[By Cor. 2 it cuts C, r, at a constant angle y. But (Art. 4 (1)) 4 sin2y=t2/p. r; therefore, etc. In the particular case when C, rз is a limiting point we have the theorem :-" if a variable circle touch two fixed circles, its radius is in a constant ratio to the square of the tangent to it from either of the limiting points." Also, "the ratio of the tangents from the limiting points is constant."]

4. A variable circle cuts two fixed circles at angles a and ẞ, tangents are drawn from its centre to the circles, and tangents t1 and to from the points of contact to the variable circle; prove that

t2/tricos a/rcos B,

and deduce the properties of Ex. 3 as particular cases (Preston). See Spherical Trigonometry, Art. 159, Ex. 15.

5. Find the locus of the centre of a circle cutting any three circles at equal or supplemental angles.

[By Cor. 4.]

6. The vertex and base of a triangle are fixed in position and the vertical angle given in magnitude; find the envelope of the circumcircle.

SECTION III.

CIRCLE OF SIMILITUDE.

94. Let A, r1; B, r2 be any two circles, Z and Z' the points of section of AB such that

then the segments AB and ZZ' divide each other harmonically, and the circle C, r, on ZZ' as diameter is termed their Circle of Similitude. The points Z and Z are the Internal and External Centres of Similitude.

95. The circle of similitude has the following fundamental properties :—

3

relation CA. CB=r,2 (Art. 70), or the centres of the given circles are inverse points with respect to their circle of similitude.

2o. The points Z and Z are the intersections of the transverse and direct common tangents.

3o. It is coaxal with the given circles.

[For Z and Zare on the same circle coaxal with A and B, since the ratios of the tangents from them are each equal to the ratio of the radii, and only one circle coaxal with A, r1 and B, r2 can contain these points, viz. that on the line ZZ' as diameter.]

4°. From Cor. 3 it is the locus of a point such that the tangents drawn from it to the circle have the constant ratio of the radii.

[Cf. Art. 88, Cor. 2.]

This follows independently, since PZ and PZ are the bisectors of the angle APB, hence

PA|PB=AZ|BZ=AR/BS;

therefore, etc., by Euc. VI. 7.

5°. The circles subtend equal angles at any point on it. (By 4°.)

2

6o. In the particular case when the circle B, r, becomes a right line the centre B is at infinity, its inverse A (Cor. 1) coincides with C, therefore the centres of similitude of a line and circle are the extremities of the diameter of the circle perpendicular to the line.

EXAMPLES.

1. The circles of similitude of any three circles taken in pairs are coaxal.

[Their centres are collinear, Art. 72, Ex. 21; therefore, etc., Art. 88, Ex. 13, 2°.]

2. A circle cuts two at angles a and B; find the angle it makes with their circle of similitude.

3. The tangents from any point P on the circle of similitude to the circles A, r, and B, r2 meet them at R and S; prove (a) the chords which the circles intercept on the line RS are equal to one another; (B) The tangents from R and S to the circles B and A are equal. [Compare Art. 89, Cor. 4.]

4. The circle on the third diagonal of a complete cyclic quadrilateral is the circle of similitude of those described on the remaining two.

[Let ABCD be the quadrilateral, LMN its diagonal line, PP' the third diagonal, BD=2r1, CA=2r2, PP' = 2r3. Join PM, PN.

The triangles PAC and PBD are similar, Euc. III. 21; hence, since PN and PM are homologous lines, PBM and PCN are similar; therefore PM/PN=r1/r2. Similarly, P'M/P'N=r/r2; therefore P and P' lie on a circle to which M and N are inverse points. Also the circles on the three diagonals are coaxal; therefore, etc. It follows also by 1° that LM. LN=r32.]

5. Having given the three diagonals of a cyclic quadrilateral; to construct it.

[Let O be the centre of the circle and 71, 72, 73 the diagonals. By Ex. 4 LM. LN=r2, and is therefore known. Also LM/LN=r2/r22; hence the lines LM and LN are determined. LM=r1rs/r2, LN=r1r2/r3, and MN: But OM and ON are known (Euc. I. 47),

=

13 7172