what has been just proved, that the perpendicular on the external axis of similitude contains the centre of a circle touching three given circles, either all externally, or all internally. If we change the sign of r, the equation of the locus which we found denotes a perpendicular on one of the other axes of similitude which will contain the centre of the circle touching S externally, and the other two internally, or vice versa. Eight circles in all can be drawn to touch three given circles, and their centres lie, a pair on each of the perpendiculars let fall from the radical centre on the four axes of similitude. *119. To describe a circle touching three given circles. We have found one locus on which the centre must lie, and we could find another by eliminating R between the two conditions S= R* + 2Rr, S' = R2 + 2Rr'. The result, however, would not represent a circle, and the solution will therefore be more elementary, if instead of seeking the coordinates of the centre of the touching circle, we look for those of its point of contact with one of the given circles. We have already one relation connecting these coordinates, since the point lies on a given circle, therefore another relation between them will suffice completely to determine the point.* Let us for simplicity take for origin the centre of the circle, the point of contact with which we are seeking, that is to say, let us take a=0, B=0, then if A and B be the coordinates of the centre of Σ, the sought circle, we have seen that they fulfil the relations S-S' = 2R (r - r'), S- S" = 2R (r — r'). But if x and y be the coordinates of the point of contact of Σ with S, we have from similar triangles Now if in the equation of any right line we substitute mx, my for x and y, the result will evidently be the same as if we multiply the whole equation by m, and subtract (m-1) times the absolute term. Hence, remembering that the absolute term in S- S' is * This solution is by M. Gergonne, Annales des Mathématiques, vol. VII. p. 289. (Art. 105) r ́3 — r2 — a” – ẞ”, the result of making the above substitutions for A and B in (S- S') = 2R (r — r') is or R+r +1 R r ( S − S') + (a” + B12 + y2 - j'2) = 2R (r — r′), (R+ r) ( S − S' ) = R {(r — r' )2 — a′′ – ß" }. #12 Similarly (R+r) ( S − S") = R {(r — r'"')2 — a''2 — ß'''}. Eliminating R, the point of contact is determined as one of the intersections of the circle S with the right line 120. To complete the geometrical solution of the problem, it is necessary to show how to construct the line whose equation has been just found. It obviously passes through the radical centre of the circles; and a second point on it is found as follows: Write at full length for S-S' (Art. 105), and the equation is 2a'x+28'y+r"" — po2 — œ22 — B12 12 = Add 1 to both sides of the equation, and we have showing that the above line passes through the intersection of a'x + B'y + (r' − r) r = 0, a′′x+B′′y + (r" − r) r = 0. But the first of these lines (Art. 113) is the chord of common tangents of the circles S and S'; or, in other words (Art. 114), is the polar with regard to S of the centre of similitude of these circles. And, in like manner, the second line is the polar of the centre of similitude of S and S"; therefore (since the intersection of any two lines is the pole of the line joining their poles) the intersection of the lines a'x + B'y + (r' − r) r = 0, a′′x + B'y + (r" − r) r = is the pole of the axis of similitude of the three circles, with regard to the circle S. Hence we obtain the following construction: Drawing any of the four axes of similitude of the three circles, take its pole with respect to each circle, and join the points so found (P, P', P'") with the radical centre; then, S s" if the joining lines meet the circles in the points (a, b; a', b'; a", b′′), S the circle through a, a', a" will S' PC R 121. It is useful to show how the preceding results may be derived without algebraical calculations. (1) By Cor., Art. 117, the lines ab, a'b', a"b" meet in a point, viz., the centre of similitude of the circles aa'a", bb'b". (2) In like manner a'a", b'b" intersect in S, the centre of similitude of C', C". (3) Hence (Art. 116) the transverse lines a'b', a"b" intersect on the radical axis of C', C". So again a"b", ab intersect on the radical axis of C", C. Therefore the point R (the centre of similitude of aa'a", bb'b") must be the radical centre of the circles C, C', C". (4) In.like manner, since a'b', a"b" pass through a centre of similitude of aa'a", bb'b"; therefore (Art. 116) a'a", b'b" meet on the radical axis of these two circles. So again the points S' and S" must lie on the same radical axis; therefore SS'S", the axis of similitude of the circles C, C', C", is the radical axis of the circles aa'a", bb'b". (5) Since a"b" passes through the centre of similitude of aa'a", bb'b", therefore (Art. 116) the tangents to these circles where it meets them intersect on the radical axis SS'S". But this point of intersection must plainly be the pole of a"b" with regard to the circle C". Now since the pole of a"b" lies on SS'S", therefore (Art. 98) the pole of SS'S" with regard to C" lies on a"b". Hence a"b" is constructed by joining the radical centre to the pole of SS'S" with regard to C". (6) Since the centre of similitude of two circles is on the line joining their centres, and the radical axis is perpendicular to that line, we learn (as in Art. 118) that the line joining the centres of aa'a", bb'b" passes through R, and is perpendicular to SS'S". 121 (a).* Dr. Casey has given a solution of the problem we are considering, depending on the following principle due to him: If four circles be all touched by the same fifth circle, the lengths of their common tangents are connected by the following relation, 12.34 + 14.23 +13.240, where 12 denotes the length of a common tangent to the first and second circles, &c. This may be proved by expressing each common tangent in terms of the length of the line joining the points where the circles touch the common touching circle. Let R be the radius of the latter circle whose centre is O, r and r' of the circles whose centres are A and B, then, from the isosceles triangle a Ob, we have ab=2R sina Ob. But from the triangle AOB, whose base is D, and sides R-r, R-r', we have a B Now the numerator of this frac of the common tangent 12, hence But since the four points of contact form a quadrilateral inscribed in a circle, its sides and diagonals are connected by the relation ab.cd+ad.bc ac.bd. Substitute in this equation the = expression just given for each chord in terms of the corresponding common tangent, and suppress the numerator R2 and the denominator √(R-r) (R − r') (R − r") (R −r") which are common to every term, and there remains the relation which we are required to prove. 121 (6). Let now the fourth circle reduce itself to a point, this will be a point on the circle touching the other three, and * In order to avoid confusion in the references, I retain the numbering of the articles in the fourth edition, and mark separately those articles which have been since added. 41, 42, 43 will denote the lengths of the tangents from that point to these three circles. But the lengths of these tangents are (Art. 90) the square roots of the results of substituting the coordinates of that point in the equations of the circles. We see then that the coordinates of any point on the circle which touches three others must fulfil the relation 23 √(S) ± 31 √(S') ± 12 √(S′′) = 0. If this equation be cleared of radicals it will be found to be one of the fourth degree, and when 23, 31, 12 are the direct common tangents, it will be the product of the equations of the two circles (see fig., p. 112) which touch either all externally or all internally. 121 (c). The principle just used may also be established without assuming the relation connecting the sides and diagonals of an inscribed quadrilateral. If on each radius vector OP to a curve we take, as in Ex. 4, p. 96, a part OQ inversely proportional to OP, the locus of Q is a curve which is called the inverse of the given curve. It is found without difficulty that the equation of the inverse of the circle x2 + y2+2gx + 2fy + c is c (x2 + y2) + 2gx + 2fy +1=0, which denotes a circle, except when c=0 (that is to say, when the point is on the circle), in which case the inverse is a right line. Conversely, the inverse of a right line is a circle passing through the point 0. Now Dr. Casey has noticed that if we are given a pair of circles, and form the inverse pair with regard to any point, then the ratio of the square of a common tangent to the product of the radii is the same for each pair of circles.* For if in g+f*- c, which (Art. 80) is r*, we 9f1 substitute for g, f, c; we find that the radius of the inverse circle is r divided by e; and if we make a similar substitution in e+e' - 2gg' - 2ff', which (Ex. 1, p. 102) is D--, we get the same quantity divided by cc'. Hence the ratio of D2 - p2 — p′′ to rr' is the same for a pair of circles 12 This is equivalent (see Ex. 8, p. 103) to saying that the angle of intersection is the same for each pair, as may easily be proved geometrically. |