and lie upon the common radical axis of the three circles having the diagonals of the tetragram as diameters. 278°. Theorem.—In general a system of co-axal circles inverts into a co-axal system of the same species. (1.) Let the circles be of the c.p.-species. The common points become two points by inversion, and the inverses of all the circles pass through them. Therefore the inverted system is one of c.p.-circles. Cor. 1. The axis of the system (LL' of Fig. to 274°) inverts into a circle through the centre of inversion (261°, Cor. 1), and as all the inverted circles cut this orthogonally, the axis of the system and the two common points invert into a circle through the centre and a pair of inverse points to it. (258°, Conv.) Cor. 2. If one of the common points be taken as the centre of inversion, its inverse is at ∞. The axis of the system then inverts into a circle through the centre of inversion, and having the inverse of the other common point as its centre, and all the circles of the system invert into centre-lines to this circle. (2.) Let the circles be of the l.p.-species. Let the circles S and S' pass through the limiting points and be thus c.p.-circles. Generally S and S' invert into circles which cut the inverses of all the other circles orthogonally. (264°) .. the intersections of the inverses of S and S' are limiting points, and the inverted system is of the l.p.-species. Cor. 3. The axis of the system (MM' of Fig. to 274°) becomes a circle through the centre and passing through the limiting points of the inverted system, thus becoming one of the c.p.-circles of the system. Cor. 4. If one of the limiting points be made the centre of inversion, the circles S and S' become centre-lines, and the l.p.-circles become concentric circles. Hence concentric circles are co-axal, their radical axis being at ∞. EXERCISES. 1. What does the radical axis of (1, 278°) become? 2. What does the radical axis of (2, 278°) become? 3. How would you invert a system of concentric circles into a common system of l.p.-circles? 4. How would you invert a pencil of rays into a system of c.p.-circles. 5. The circles of 277° may be c.p. or l.p.-circles. 279°. Theorem.-Any two circles can be inverted into equal circles. But, since P and Q and also P' and Q' are inverse points, and (275°, Cor. 5) O lies on a circle co-axal with S and S'. And with any point on this circle as a centre of inversion S and S' invert into equal circles. Cor. 1. Any three non-co-axal circles can be inverted into equal circles. For, let the circles be S, S', S”, and let Z denote the locus of O for which S and S' invert into equal circles, and Z' the locus of O for which S and S" invert into equal circles. Then Z and Z' are circles of which Z is co-axal with S and S', and Z' is co-axal with S and S". And, as S, S', and S" are not co-axal, Z and Z' intersect in two points, with either of which as centre of inversion the three given circles can be inverted into equal circles. Cor. 2. If S, S', and S" be l.p.-circles, Z and Z' being co-axal with them cannot intersect, and no centre exists with which the three given circles can be inverted into equal circles. But if S, S' and S" be c p.-circles, Z and Z' intersect in the common points, and the given circles invert into line circles with centres at infinity, and having each an infinite radius these circles may be considered as being equal. (278°, Cor. 2) Cor. 3. In general a circle can readily be found to touch three equal circles. Hence by inverting a system of three circles into equal circles, drawing a circle to touch the three, and then re-inverting we obtain a circle which touches three given circles. If the three circles are co-axal, no circle can be found to touch the three. 280°. Let the circles S and S', with centres A and B and radiir and r, be cut by the circle Z with centre at O and radius OP R. Let NL be the radical axis of S and S'. Since AP is to the tangent at P to the circle S, and OP is to the tangent at P to the circle Z, .. the LAPO=0 is the angle of intersection of the circles S and Z (115°, Def. 1). Similarly BQO= is the angle of intersection of the circles S' and Z. Now and ... But OP' OQ' 2(r'cos-rcos @). R. OP' - R. OQ′=OT2-OT” (where OT is the tangent from O to S, etc.) = 2AB. OL, (275°) Cor. 1. When 0 and $ are constant, R varies as OL. .. a variable circle which cuts two circles at constant angles has its radius varying as the distance of its centre from the radical axis of the circles. Cor. 2. Under the conditions of Cor. 1 ON varies as OL, .. a variable circle which cuts two circles at a constant angle cuts their radical axis at a constant angle. Cor. 3. When OL=o, r'cos p=r cos 0, and r: rcos: cos 0. ... a circle with its centre on the radical axis of two other circles cuts them at angles whose cosines are inversely as the radii of the circles. Cor. 4. If circle Z touches S and S', 0 and 4 are both zero or both equal to π, or one is zero and the other is π. .. when Z touches S and S', R= AB OL, where the variation in sign gives the four possible varieties of contact. Cor. 5. When 0=4=22, Z cuts S and S' orthogonally, and OL=0, and the centre of the cutting circle is on the radical axis of the two. SECTION VII. CENTRE AND AXES OF SIMILITUDE OR The relations of two triangles in perspective have been given in Art. 254°. We here propose to extend these relations to the polygon and the circle. 281°. Let O, any point, be connected with the vertices A, B, C, of a polygon, and on B X x с A ... OA, OB, OC, ... let points a, b, ... OA: Oa=OB: 06=OC: Oc... and OA: Oa=OB: OOC: Oc'... Then, since OAB is a ▲ and ab is so drawn as to divide the sides a' proportionally in the same order, .. ab is to AB. (202°, Conv.) Similarly, bc is || to BC, cd to CD, etc., similarly, and b'd is to BC, c'd' to CD, etc., AOABAOab Od'b', AOBC Obcob'c', ... .. the polygons ABC..., abc..., and a'b'c'... are all similar and have their homologous sides parallel. Def. The polygons ABCD... and abcd... are said to be similarly placed, and O is their external centre of similitude; while the polygons ABCD... and a'b'c'd'... are oppositely placed, and O is their internal centre of similitude. Hence, when the lines joining any point to the vertices of a polygon are all divided in the same manner and in the same order, the points of division are the vertices of a second |