parallel to the latter, and the line of centres is the diameter at right angles to L; therefore, etc.

110. It has been seen as a particular case of a general property of coaxal circles (Art. 93) that any line A,Ã1⁄2Ð ̧Ð1⁄2 through C, a°, cuts the circles at equal angles and, B°, that the intercepted chords A,4, and BB, are in the ratio of the radii. These are obvious by the following method:

2

Join AA, and BB1. Since CA/CB=r/r2 = AA,/BB, the triangles CAA, and CBB, are similar (Euc. VI. 7); therefore AA, is parallel to BB1, and similarly AA2 to BB2 Hence the isosceles triangles AAA, and BÂ ̧Â1⁄2 are similar, whence, a, the angles A,AA, and BBB are equal, and, B°, A1A2/B1B2=rı/r2

Definitions. A, and B, are termed Homologous Points; and since the radii AA, and BB, through them are parallel, the tangents at homologous points on the circles are

n

n

2

parallel. Thus the tangents at A, and B, are parallel. More generally any two points A and B, which connect through C such that CA2/CB2 = r1/r2 are homologous. A and B, are termed Antihomologous Points, and since the radii AA, and BB, through them make equal angles with their line of connexion, the tangents at antihomologous points meet on the radical axis.

41

4

Let a second transversal through C meet the circles in AABВ The chords Д12 and BB, joining pairs of homologous points are termed Homologous Lines, and those joining pairs of antihomologous points Antihomologous Lines. Thus  ÂÎВ and  ̧à ¿ are pairs of antihomologous lines.

4

3,

4

111. Theorem.-Homologous chords (Â ̧Â ̧, В ̧Â ̧) of

any two circles are parallel.

For it has been shown that AA, and BB1, AA, and BB2 are pairs of parallel lines; hence the two isosceles triangles ÂÂÂ。 and BBB have equal vertical angles, and are therefore similar (Euc. VI. 6).

3

NOTE. Since any line through C meets homologous lines AA, and BB, in homologous points A, and B2, therefore An, B2 are in general the corresponding intersections of pairs of homologous lines. The two points  ̧ ̧  and BB, BВ are homologous.

3'

4

112. Theorem.-Antihomologous chords (AA, BÎÂ1⁄2) of any two circles meet on their radical axis.

=

=

By Art. 111, we have CA/CA,= CB/CB2, but (Euc. III. 36) CA/CA, CA/CA2; hence CB1/CB, CA/CA, or CA2. CB1 = CA. CB ̧; thus :--any two points are concyclic with the corresponding pair of antihomologous points; therefore, etc. (Art. 88, Ex. 6).

PRODUCTS OF ANTISIMILITUDE.

113. By the previous Article, we have from the cyclic quadrilateral A‚ ̧ ̧

CA2. CB1 = CA · CB3.

We may therefore infer that the rectangle under the distances of either centre of similitude from a pair of antihomologous points is constant.

If the circles A, î1; B, r2 be regarded as portions of two geometrical figures, any point An of one is antihomologous to Bn of the other when the line AnBn passes through a centre of similitude C, and CA, . CB2 is equal to the above constant, which is termed the Product of Antisimilitude (External or Internal).

To find the values of the products, we take the extreme positions of the variable line CAB1 which for real intersections are the common tangents.

We have therefore

CA2. CB1=CT1. CT2.............

2

2'

.(1)

Again, since TT, subtends a right angle at each of the

2

limiting points M and N (Art. 88, Cor. 3),

CT1. CT2 = CM. CN.....

2

These constant values which may be expressed in terms of the distance (8) between the centres of the given circles and their radii (r, and r2) are of importance in the theory of coaxal circles, and will frequently be made use of in the next chapter.

1

Similarly the internal product of antisimilitude is found to be equal to

2

NOTE.-It should be noticed when the two circles lie wholly outside each other 8>r1+r, if they intersect &<r1+r1⁄2 and > T1 ~ r2 (Euc. I. 20), and when one lies completely within the other 8<r1 ~ r2 (Euc. III. 12); hence it follows from (3) that the external product of antisimilitude is negative only when one circle lies wholly within the other. Also from (4) the internal product is negative when the circles are external to one another and positive in every other case. In the case where both products are positive &>r1 ~ r2 and <r1+r2; therefore 8, 71, 1⁄2 form a triangle (Euc. I. 20), or the circles intersect in a pair of real points.

EXAMPLES.

1. If a variable circle touch two circles with contacts of similar species, its points of contact are antihomologous points.

[By Art. 112, if AA, and BB1 be produced to meet in X, XB1=XA1⁄2. In the case of internal contact the points of contact are A1, B2.]

2. Describe a circle passing through a given point (P) and touching two fixed circles (A, rı) (B, r2).

[By Art. 110, the required circle passes through an antihomologous point P', and the problem thus reduces to "describe a circle passing through two fixed points and touching a given circle."]

3. The polars of the external centre of similitude with respect to two circles are equidistant from the radical axis, and therefore also from the limiting points.

4. The line at infinity is an axis of perspective of two circles. [Regard the circles as similar polygons of an infinite number of sides, and join their corresponding vertices (ie. the homologous points). Thus the ex-centre of similitude is a Centre of Perspective of the circles. Again, the corresponding sides (i.e. homologous lines) intersect on the axis of perspective. In this case they are parallel.

Hence the line at infinity is the axis of perspective of every two circles. (Cf. Art. 87).]

5. The radical axis is also an axis of perspective of two circles. [For since antihomologous points B1, A2 connect through a centre of similitude C, the circles may be regarded as polygons of an infinite number of sides whose corresponding vertices are antihomologous points and whose corresponding sides are therefore antihomologous lines; but these latter intersect on the radical axis (Art. 112), which is therefore the axis of perspective.*

6. The poles An, B2 of the chords A12 and B1B2 are homologous points.

[For they are the intersections of pairs of homologous lines, viz. the tangents at A1, A2 and B1, B2 respectively.]

7. In Ex. 6 the lines A,B and AnBn are conjugate with respect to both circles.

8. If C, C' denote the centres of similitude of two circles which cut orthogonally at X; the inverse (C") of the point C' with respect to the circle A is the inverse of C with respect to the circle B.

[Since C' and C" are inverse points, AC"X-AXC'=45°; hence AC"X=BXC, therefore CB/BX=BX/BC", therefore etc.

9. A variable circle touches two equal circles with contacts of opposite species: show that the product of the intercepts on their transverse common tangents made by the perpendiculars from the centre and measured from their point of intersection is constant.

10. The centres of similitude, the centre of the circle of similitude, and the centre of either circle B are pairs of inverse points with respect to a circle concentric with A.

* Two circles are thus shown to be doubly in perspective to each centre of similitude; the two axes of perspective forming the coaxal circle whose radius is infinitely great, viz., the radical axis and the line at infinity. It follows that "for every two circles in the same plane, however circumstanced as to magnitude and position, the radical axis and the line at infinity, being both axes of perspective, are both chords of intersection; the corresponding points of intersection, real or imaginary, according to circumstances in the case of the former, being of course from the nature of the figures always imaginary in the case of the latter." (Townsend.)