11. The poles A1, B1 of the radical axis of two circles (A, r1; B, r2) are inverse points with respect to their circle of similitude.

[For since

also since

By addition

AA1. AL=r2, angle APL-A A1P;
BB1. BL=r,2, angle BPL- BB1P.
APB PAB1+PB1A1=π- A1PB1.

Thus A1, B1 and A, B, since they subtend similar angles at P, are pairs of inverse points with respect to the circle of similitude (Art. 72, Cor. 8).]

12. If a variable circle V cut two circles A and B at constant angles, show that the centre of similitude of any two positions Vi and V2 is on L the radical axis of A and B.

[For V1 and V2 meet the line L at equal angles (Art. 88, Ex. 7); therefore it passes through their ex-centre of similitude.]

12a. Hence show that if the circles A and B each cut three fixed circles V1, V2, V3 at the same angles a, ẞ, y, an axis of similitude of the three is the radical axis of the two.

13. Construct a quadrilateral, having given the four sides, and that two adjacent angles are equal. (Mathesis, 1881.)

14. Feuerbach's Theorem. To prove by an method that the nine-points-circle touches the in-circle.

elementary

Draw C'X the fourth common tangent to the in- and ex-circles to the side c of the triangle ABC. We shall prove that the line joining M, the middle point of the base, to the point of contact X passes through the point of contact I of the in- and nine-pointscircles.

Let T be the point of contact of the in-circle, P the foot of the perpendicular, and C' the foot of the internal bisector of C.

P

By Art. 71, Ex. 3, MP. MC' = (ab)2= MT2 = MX. MY. Hence XYPC' is a cyclic quadrilateral and angle MC'X=MYP; but

MC'X= MC'C-XC'C=AB; hence MYP AB, and therefore Y is on the nine-points-circle, since the latter cuts the base AB at this angle. Therefore the circles cut or touch at Y. But the tangents at M and X to the circles are parallel, since they both meet the base at the same angle AB. M and X are thus homologous points.

15. The straight lines joining the points of contact of the fourth common tangents to the in- and three ex-circles to the middle points of the corresponding sides are concurrent. (Dublin Univ. Exam. Papers.)

[By Ex. 14, the point of concurrence is where the nine-pointstouches the in-circle.]

16. A right line ABCD is drawn across two circles cutting them at angles a and ẞ respectively; show that if a variable circle cuts the given ones at the same angles in the points A', B', C', D', AA', BB', CC', DD' are concurrent; and find the locus of their point of

concurrence.

[The given circles meet the line ABCD and circle A'B'C'D' at equal angles; hence AA' etc. are antihomologous points with respect to the external centre of similitude of the latter. Therefore AA' etc. meet on the circle A'B'C'D' at a point (P) the tangent at which is parallel to ABCD. The locus of P is the radical axis of the fixed

circles by Ex. 12.]

SECTION II.

CIRCLES OF ANTISIMILITUDE.

Definitions. The circle described with either centre of similitude of two given circles as centre, the square of whose radius is equal to the corresponding product (Art. 113) of antisimilitude, is known as a Circle of Antisimilitude.

Thus there are two circles of antisimilitude, External and Internal, according as the centre coincides with the external or internal centre of similitude of the given circles.

From the definition it is evident that all pairs of antihomologous points are inverse points with respect to the circle of antisimilitude, or, more generally, that each of the two given circles is the inverse of the other with respect to either circle of antisimilitude.

In the next chapter this latter circle, from this fundamental property, will be otherwise known as the Circle of Inversion of the two given ones.

114. The following theorems are of importance in the geometry of these circles.

1o. Any two circles A and B and their circles of antisimilitude are coaxal.

For the constant product CA2. CB1 (Art. 113) has been proved equal to CM. CN; hence M and N are a common pair of inverse points to the four circles.

2o. The squares of the tangents t1 and t2 from any point of either circle of antisimilitude to A and B are in the ratio of the radii; or t2: t2 = r1: 12

Since the circles are coaxal,

=

t2t22 CA: CB = r12 (Art. 88, Cor. 2.)

3°. The external circle of antisimilitude cuts orthogonally all circles cutting A and B at equal angles.

Since AA, and BB1 are equally inclined to the line AB1, if they are produced to meet in X, then XB1A, is an isosceles triangle, and X is therefore the centre of a circle cutting A and B at equal angles.

Thus any circle cutting A and B at equal angles passes through a pair of inverse points A, and B, with respect to the ex-circle of antisimilitude; therefore, etc.

2

See also the method of Art. 93, Cors. 3, 4.

4°. Any circle intersecting A and B at supplemental angles is orthogonal to the internal circle of antisimilitude.

[Proof similar to 3°.]

5°. Any circle intersecting A and B orthogonally is orthogonal to both their circles of antisimilitude.

For in this particular case A and B are cut at angles which are at once both equal and supplemental; there fore, etc. by 3° and 4° combined.

EXAMPLES.

1. A variable circle passing through a fixed point and cutting two given ones at equal angles passes through a second fixed point. [In every position it passes through the inverse of the fixed point with respect to the ex-circle of antisimilitude.]

2. A variable circle passing through a fixed point and cutting two fixed circles at supplemental angles passes through a second fixed point.

[The inverse of the given one with respect to the in-circle of antisimilitude.]

3. Two circles X, Y intersecting two others A and B at equal angles have for radical axis a line passing through the centre C of the ex-circle of antisimilitude of A and B.

[For if X and Y intersect in a point P, each must pass through the inverse of P with respect to C.]

3a. If the angles are supplemental, the radical axis of X and Y passes through the in-centre of antisimilitude.

4. If three circles X, Y, Z meet two others A and B at equal or supplemental angles, the radical centre of the three coincides with the external or internal centre of similitude C or C' of the two.

[For by Ex. 3 the radical axes of Y, Z; Z, X; X, Y each pass through Cor C" according as the angles of section are equal or supplemental; therefore, etc.]

NOTE. In this example it may be noticed that in the first case the circles A and B each cut X, Y, and Z at equal angles; therefore they cut the ex-circles of antisimilitude of Y, Z; Z, X; X, Yat right angles (Art. 114). But the ex-circles of antisimilitude are coaxal; hence a variable circle A cutting three others X, Y, Z at equal angles describes a coaxal system, the conjugate of that formed by the circles of antisimilitude of X, Y, Z taken two and two. More generally, a variable circle cutting three others X, Y, Z at similar angles describes four coaxal systems whose radical axes are the four axes of similitude of X, Y, Z. Also, since the common ortho