23. Let the bisector of the LPCA meet PQ at R. Join RA. Then by 1. 4, the AS CPR, CAR are identically equal; :: _RAC is a rt. _ ; hence RA is the tangent to both os at A [111. 16). Thus the bisector of the < PCA meets PQ at the point at which it is cut by the tangent at A. Similarly the bisector of the LQC'A meets PQ at the same point: that is, the bisectors intersect on PQ; and are at rt. angles, for they are also the bisectors of the ¿ PRA, QRA [Ex. 2, p. 29]. 24. Let o, c' be the centres of the two os. From centre C with radius equal to the difference of the radii of the given O $, describe a o to cut cc' at X, Y; and from C' draw the tangent c'p'. Then [Ex. 17, p. 218] PQ = C'P. .. the sq. on PQ= the sq. on p'c' the rect. C'X, C'Y 25. Let A be the centre of the o to which the tangent is to be drawn, and B the centre of the which is to cut off from the tangent an intercept equal to K. In the O (8) place a chord equal to K, and describe a concentric o to touch this chord (i.e. to pass through its middle point). Then draw a common tangent to the O (A) and the o of construction. Then the O (B) will cut off from this tangent a part equal to k [Ex. 5, p. 181]. Impossible when K is greater than the diameter of the O (B), or when, of the circle (A) and the o of construction, one falls within the other. In general there are four solutions. 26. Let A and B be the centres of the given O$, H and k the two given lines. Place chords equal to H and K respectively in the OS (A) and (B), and describe concentric os touching these chords. Then draw a common tangent to the two of of construction. From this tangent the two given of will cut off parts equal to H and K [Ex. 5, p. 181]. PROBLEMS ON TANGENCY, Page 220. Loci. (i) The st. line which bisects the line joining the given points at rt. angles. (ii) The st. line perp. to the given st. line at the given point. (iii) The radius through the given point, indefinitely produced both ways. (iv) Two st. lines part . to the given line, one on each side of it, at a perp. distance from it equal to the radius of the touching circles. (v) Two concentric circles, whose radii are ra+r, and ri~1 22 where ri is the radius of the given circle, and r, the radius of the circles which touch it externally or internally. (vi) The two st. lines which bisect internally and externally the angle between the two given st. lines. 27. The three given st. lines are supposed to be of infinite length. The locus of the centres of Os touching any pair must be the internal and external bisectors of the angle between them. Four different centres will be given by the intersection of these loci, corresponding to what are known as the inscribed and escribed os of the formed by the three given lines. 28. Let AB be the given st. line, c the given point in it, and D the other point through which the required is to pass. Then since the required o is to touch AB at C, its centre must lie on the st. line through c perp. to AB. Again, since the required o is to pass both through C and D, its centre lies on the st. line which bisects CD at rt. angles. Therefore o, the intersection of these loci, is the centre of the required o. One solution: except when D is in AB, then impossible, for the loci will in that case never meet. 29. Let c be the centre of the given 0, A the point on its oc, and D the other point through which the required o is to pass. Then since the required o is to touch the given O at A, ..its centre must lie on CA, or CA produced (111. 11, 12]. Again, since the required o is to pass through the points A and D, its centre must lie on the st. line which bisects AD at rt. angles. .. o, the intersection of these loci, is the centre of the required o. One solution : except when D lies on the tangent at A; then impossible, for the loci in that case will never meet. 30. Let y be the given radius, AB the given st. line, c the given point. (i) Then since the required o is to touch AB, its centre must lie on one or other of the two st. lines par. to AB and at a distance from it equal to r. (ii) Again, since the required o is to pass through c, its centre must lie on the oce of a o of which c is the centre, and r the radius. Hence the intersections of either st. line in (i) with the o in (ii) will give centres of the required o. Theoretically there will be four solutions. (i) If c is in AB, the circle-locus will touch both of the parls, and there will be two pairs of coincident solutions. (ii) If c is not in AB, the circle-locus can only cut that parallel which is on the same side of AB as C: thus of the four theoretical solutions, two will be impossible, and the other two will be distinct, coincident or impossible as the distance of o from AB is less, equal to, or greater than 2r. 31. Let A and B be the centres of the given Os, and r1, rz their radij; and let r be the radius of the required circle. (i) Then the centres of all os of radius r which touch theo (A), lie on one or other of the concentric Os whose radii are ra +r, or ra~ g respectively. (ii) Again, the centres of all os of radius r which touch the 0 (B), lie on one or other of the concentric OS whose radii are r2 + p or r2 ~r. Hence the intersections of either o in (i) with either o in (ii) give centres of the required o. Thus theoretically we get eight solutions. Which of them are real, and which impossible will be found to depend upon the relative magnitudes of ru, ro, and and also upon the 32. relative position of the two given 08 - whether one is without the other, one within the other, or whether they intersect. Let AB, CD be the two given st. lines, and r the radius of the req. O. (i) Then all os of radius r which touch AB must have their centres on one or other of the st. lines par!. to AB, and at a perp. distance from it equal to r. (ii) Similarly all os of radius p which touch BC must have their centres on one or other of the st. lines par to BC, and at a perp. distance from it equal to r. Hence the intersections of either st. line in (i) with either st. line in (ii) gives a centre of the required O. Thus there will be four solutions, all of which will be real, when the given lines intersect. If AB and CD are par'., the method fails. In this case there will be no real solution, unless r= half the perp. distance between AB and CD: then there will be an infinite number of solutions. 33. Let AB be the given st. line, r, the radius of the given 0,. the radius of the required O. (i) Then the centres of all os of radius r, which touch the given 0, will lie on a concentric o of radius 9, +r or ri~r. (ii) And the centres of all os of radius r, which touch AB, will lie on one or other of the st. lines parl. to AB at a distance from it equal to r. Hence the intersections of either o in (i) with either st. line in (ii) give centres of the required o. Thus theoretically there are eight solutions. Let a denote the distance of AB from the given centre. Then if x is greater than rı + 2r all the solutions are impossible. If x=r, + 2r then two solutions are coincident, the rest impossible. If lie between r, and r, + 2r, two solutions are real (and distinct), the rest impossible. Again, if x=r,, there are two 74 KEY TO EUCLID. rual pairs of coincident solutions, the rest impossibla; and if ac is less than r six solutions are possible. Finally, all eight solutions are possible, if rı > 2r and <r; – 2r. 34. Let AB be the given st. line, and ru, r, the radii of the given 0% Describe a o of radius r, to touch AB. (i) Then the centre of 2nd required circle must lie on one or other of the concentric Os whose radii are ri+r, or r~rg. (ii) The centre of the 2nd required o must also lie on the st. line par. to AB at a distance from it equal to ra, and on the same side of it as the 1st O. Hence theoretically we have four solutions. The 0, whose radius is ra +r2, will always give two possible distinct solutions. The o whose radius is ra~r2 gives two coincident solutions. 35. Let PQ be the given line, and c the centre of the given O ; and let a second o, whose centre is F, touch the given o at E and PQ at A. Then shall AE produced meet the o ce of the given o at D, an extremity of the diameter perp. to PQ. Join DC, FA, CF. Then CF passes through E [111. 12]. Now LFAE = L FEA, because FA = =FE; = L CED [1. 15] =LCDE, because CD=CE. :. DC is par!. to FA; but FA is perp. to PQ [111. 18]. :. AE passes through an extremity of the diameter perp. to QP. 36. Because CD= CE, .. LCDE: = L CED; =- FEA (1. 15]. Also LCDE = alt. - FAE (1. 29]. i. _FEA = L FAE; .. FE= FA. Now FE produced passes through the centre C, and FA is perp. to PQ; a o described from centre F with radius FA satisfies the required conditions (11. 12 and 16]. |