touch one, two, or three given circles, or which shall touch both a straight line and a circle, or two straight lines and a circle, or a straight line and two circles, and at the same time pass through one or two given points, as the other data may happen to admit. The six new problems of contact which are thus suggested are too remarkable to be passed over without further notice; they are accordingly here subjoined. 1. To describe a circle which shall pass through two given points A, B, and touch a given circle C D E. G E In the circumference CDE take any point C, and describe (59.) a circle which may pass through the three points A, B, C; then, if this circle meet the circumference CDE in no other point, it is the circle required but if it do, let that other point be D: join A B, CD, and let them be produced to meet in F: from F draw FG, touching the circle CDE (56.): let G be the point of contact, and F B describe a circle (59.) passing through the three points A, B, G. Then, because the chord CD of the circle CDE meets the tangent GF in F (21.), the square of G F is equal to the rectangle CF, FD, that is, to the rectangle AF, FB (20.): therefore (21. Cor.) GF touches also the circle ABG, and, consequently, the circle ABG touches the circle C D E (2. Cor. 2 and 9.), and is the circle required. If A B and C D be parallels (which will be the case when the line which bisects A B at right angles passes through the centre of the circle CDE), a tangent FG is to be drawn parallel to AB or CD (57.), and the circle ABG, being then described as before, will be the circle required. It is evident, in each case, that there are two tangents FG, and two circles ABG corresponding to them, one touching the given circle externally, the other internally. Cor. The problem which requires a circle to be described which shall touch a given straight line in a given point, and also a given circle, may be solved after a similar manner; viz. by describing a circle which shall touch the given straight line in the given point, and likewise pass through a point assumed in the circumference of the circle, and then proceeding as in the proposition. 2. To describe a circle which shall pass through a given point A, and touch two given circles BC D, E F G. Suppose that the required circle is described, and that it touches the given circles in the points D, F respectively: join D F, and since the straight line DF cannot touch the given circles in the points D, F, (because then it would touch the required circle ADF in the same points, which (1. Cor. 2.) is absurd) let it be produced both ways to meet the circumferences a second time in the points C, G respectively: take H, K, the centres of the circles B CD, E F G, and join HK, HD, K F, KG: then, because the circle A D F touches BCD in D, and EFG in F, the radii HD, KF produced (8. Cor. 1.) will meet in its centre, L; and, because KG F, LFD are isosceles triangles, the angle KG F is (I. 6.) equal to K F G, that is (I. 3.) to LFD, that is again (I. 6.) to L DF or (I. 3.) HD C: therefore (I. 15.) KG is parallel to HD, and, consequently, if the circles BCD, EFG be equal, HK, DG will be parallel (I. 21.), or if one of them, as B CD, be greater than the other, H K and DG will meet, if produced in some point M. In the latter case, draw MB touching the circle BCD in B (56.), join H B, and from K draw KI (I. 45.) perpendicular to M B, and therefore (I. 14.) parallel to HB: then, because KI, H B are parallel, KI: HB::KM: HM (II.30. Cor. 2.) that is, :: KG: HD, because KG, HD are parallel: but HB is equal to HD; therefore (II. 18.) KI is equal to K G, and I is a point in the circumference of the circle E FG, and (2.) MB touches the circle EFG in the point I. Now, MI MB:: MK: MH (II. 29.), that is, :: MG: MD; therefore, alternando (II.19.), MI: MG MB: MD: but, because MI touches the circle EFG, the square of MI is equal to the rectangle M FX MG (21.), and consequently (II. 38. Cor. 1.) M F: MI:: MI: M G: therefore (II. 12.) MF: MI:: MB: MD, and (38.) the rectangle MIx M B is equal to the rectangle M F x MD. Join MA, and let it cut the circumference of the circle AFD a second time in the point N: then the rectangle MA × MN is equal (20.) to the rectangle MFx MD, that is, by what has been just demonstrated, to the rectangle MIx M B. Therefore, reversely, to solve the problem, draw the common tangent BI to the two given circles (58.), and produce it to meet the line HK which joins their centres in M: join MA: take M N a fourth proportional (II. 53.) to MA, MB, MI, and, as in (1), describe the circle AND, passing through the two points A, N and touching the circle BCD: AND will be the circle required. For if MD be joined, cutting the circle EFG in F and G, and if KF and KG be joined, the circle described will (20. Cor.) pass through F, because (as has been shown above) MDX MF is equal to MBX M I, that is to MA× MÑ; and (which has been likewise shown) K G will be parallel to HD. Therefore, if in HD produced (which (8. Cor. 1.) passes through the centre of the circle AND, because it passes through the centre H of BCD, which touches AND in D), there be taken L, the centre of the circle AND, and L F be joined, the angle LFD will be equal (I. 6.) to LDF, that is (I. 3.) to HD C, that is, again (I. 15.) to KGF, or (I. 6.) KFG; and because DFG is a straight line, K F and F L likewise lie in a straight line (I. 3.), and therefore the circle AND touches the circle EFG in F (9.). Should the given circles be equal, or should the point N coincide with the point A, the foregoing solution must be modified accordingly. The latter case is provided for by the corollary of the preceding problem: both, indeed, are comparatively easy, and are therefore left to the student. It remains to observe, that we have only considered the case in which the required circle touches both of given circles externally: since both, however, may also be touched internally, or one -internally and the other externally by the same circle, there are evidently four different circles which satisfy the conditions of the problem. For each of these a different construction is re 117 dius of C; and, as in the last problem, describe a circle D E F which shall pass through the centre F of the circle C, and touch these two new circles in the points D, E respectively: then, if a circle be described concentric with the circle D E F, with a radius less than that of DEF by the radius of C, it will evidently touch each of the three given circles (10. Cor. 2.), and will be the circle required. quired circle touches each of the given In the case here considered, the recircles externally: since, however, a circle may be described which shall touch all three internally, or any two externally, and the third internally, or again, any two internally and the third externally, there are no less than eight different circles which satisfy the problem: the construction of each of these is obtained after a similar manner. 4. To describe a circle which shall pass through a given point A, touch a given straight line B C, and also touch a given circle D E F. Take G the centre (55.) of the circle DEF, and through G draw (I. 45.) the diameter D F perpendicular to BC to meet B C in C: join A D, and take DH a fourth proportional (II. 53.) to AD, D C and D F, i. e. (II. 38.) such that ADXDH may be equal to DCX DF; and describe a circle H AB (59.) passing through the two points A, H, and touching the straight line BC: this shall be the circle required. For, let it touch B C in B; join BD, and let it cut the circumference DEF in K, and join K F: then, because (15. Cor. 1.) DK F is a right angle, the triangles D KF, DCB are equiangular, and (II. 31.) DK: DF::DC: DB, or (38.) D KXDB is equal to DCXDF, that is, to ADxDH; therefore K is also a point in the circumference of the circle A B H (20. Cor.). Now take L the centre of the circle A BH (55.) and join LB, LK, GK: and, because LB is (2. Cor. 1.) perpendicular to BC, it is (I. 14.) parallel to D C, and therefore (I. 15.) the angle LBK is equal to GDK; but LK B is equal to LBK, and GKD to GDK, because LB is equal to L K, and GD to G K (I. 6.) ; therefore the angle LKB is equal to GK D, and (I. 3.) LKG is a straight line: therefore the circle ABH touches (9.) the circle DEF in the point K, and is the circle required. If DH be equal to DA, a circle ABH is to be described touching DA in A, and also touching the straight line BC. (See the beginning of this Scholium.) In this problem there are two circles satisfying the conditions: the construction of that which touches the given circle internally will readily be understood by applying what has been said to the subjoined figure. F circle (55.) and (59.) describe the HLN passing through the point N and touching the straight lines H K, LM: then, if a circle be described concentric with HLN, and with a radius less than that of HLN by the radius of E FG, it will evidently touch the straight lines A B, and CD, and the circle E F G (10. Cor. 2.) and will therefore be the circle required. Four different circles may be described to satisfy this problem, viz. two touching the given circle externally, and two internally: the latter two may be found by drawing H K and L M between A B and CD. 6. To describe a circle which shall touch a given straight line A B, and two given circles CDE, FGH. If the given circles be equal to one another, draw K L parallel to A B, at a distance from it equal to their common radius (I. 48.), and (59.) describe a circle passing through the centres of the two given circles and touching the line K L: then it is evident that a circle concentric with the latter, and having a radius less by the common radius of the given circles, will be the circle required. But, if one of them, as FG H, be greater than the other, de K scribe about its centre a circle f g h with a radius less than that of FG H by the radius of the lesser circle CDE: draw K L parallel to A B at a distance from it equal to the radius of CDE; and describe a circle passing through the centre of CDE touching the line KL, and touching also the circle fg h, as shown in the last problem but one: then, if a circle be described concentric with this and with a radius less by the radius of CD E, it will evidently be the circle required. Here, also, there are four different circles satisfying the conditions of the problem; and the same principle leads to the construction of them all. *External contact is where two circles touch one another, and each of them is without the other; internal contact, where two circles touch one another, and one of them is within the other. In the latter case, it is evident that the circumference of the larger circle is without the circumference of the other; and yet each of them is properly said to touch the other internally, meaning, with that description of contact which is called internal, Much as each of the foregoing problems may be varied in appearance by the consideration of internal contact, they admit of being varied yet further by supposing, when a point is given, that it is in the given tangent or circumference: and when it is besides suggested that, in each of the ten problems furnished by the proposition and scholium, one of the data be changed for "a given radius," or for "a given line in which the centre is to lie," or again, two of the data for both of these, the student will have before him a field no less pleasing than extensive, still remaining for the exercise of his ingenuity, in problems of contact. PROP. 60. Prob. 7. (Euc. iii. 33.) Upon a given straight line AB to describe a segment of a circle which shall contain an angle equal to a given angle A B C. Let A QB be the segment required, and Pits centre. Then, because the angle ABC is equal to the angle in the alternate segment, the line BC touches the circle (17. Cor.) at B. Therefore, PB joined is at right angles to B C. Again, because P is the centre of a circle having the chord A B, it is in the straight line which bisects A B at right angles (3 Cor. 2). There- A fore, reversely, draw BD perpendicular to BC (I. 44.), and bisect AB at right angles A (1.43.Cor.) by a straight line, cutting BD in P: D P B with the centre P and radius PB describe the circular arc B QA: and the segment B QA will be the segment required. PROP. 61. Prob. 8. (Euc. iii. 34.) Given a point A in the circumference of a circle A B C; to cut off a segment which shall contain a given angle, by a straight line drawn from the point A. From A draw AD touching the circle, and let A B be the line required: then, because the angle BAD (17.) is equal to the angle in the alternate segment BCA, it is equal to the given angle. Therefore, reversely, make DAB equal to the given angle (I. 47); and AB will be the line required. B PROP. 62. Prob. 9. (Euc. iv. 2, 3.) Given a circle A B C, to (1) inscribe in it, or (2) describe about it, a triangle similar to a given triangle D E F. 1. Let A B C be the required inscribed triangle, and through à draw the line GAH touching the circle (56.): then the angle GA B is equal to the angle at C, and the angle H A C to the angle at B (17.) Therefore, reversely, take any point A of the circumference, draw the tangent GAH (56.), and make the angles GAB, HAC equal to the angles at F and E respectively (I. 47.) Then, if A B, A C meet the circumference in be joined, ABC will be the triangle the points B, C respectively, and if B C required; for, the angles at B and C being equal to the angles at E and F, each to each, the third angles at A and D are likewise equal (I. 19. Cor. 1.), and therefore the triangles are similar (II. 31. Cor. 1.). 2. Let KLM be the required circumscribed triangle, and let its sides touch the circle in the points A, B, C: take O the centre of the circle (55.), and join ОА, ОВ, О С. D K M Then, because the angles at A and B of the quadrilateral ALBO are right angles (2. Cor. 1.), the remaining angles L and AO B are together equal to two right angles (I. 20. Cor.) and AOB is the supplement of the angle L, or, of its equal, the angle E; which supplement may be obtained by producing the side E F. In like manner it may be shown, that the angle BO C is the supplement of the angle M or F. Therefore, reversely, taking any point A'in the circumference, at the point O, make (I. 47.) the angle A O B equal to the supplement of E, and the angle BOC equal to the supplement of F: at the points A, B, C draw tangents (56.) meeting one another in K, L, M; and KLM will be the triangle required. PROP. 63. Prob. 10. (Euc. iv. 6, 7, 11, 12, 15, 16.) A circle being given; to inscribe in it, or describe about it, 1. an equilateral triangle; or 3. a regular pentagon; or Take C the centre of the circle; and, 1. to inscribe an equilateral triangle: from A, any point of the circumference, with the radius A C, describe a circle cutting the given circle in the points B, D: produce AC to meet the circumference in E, and join B D, DE, EB. Then joining BC, CD, BA, AD, because the triangles ACB, ACD are equilateral, the angles BCA, DCA are each of them a third of two right angles (I. 6. and I. 19.). Therefore the adjacent angles BCE, DCE are each of them two-thirds of the same, as is also the angle B CD; and because the sides of the triangle BDE subtend equal angles at the centre C, they are equal to one another (12. Cor.1.), i.e. the triangle B D E is equilateral. 2. To inscribe a square draw two diameters at right angles to one another, and join their extremities: the included figure will be a square; for its sides are equal, because they subtend equal angles at the centre C (12. Cor. 1.), and its angles are right angles, because they are contained in semicircles (15. Cor. 1.) 3. To inscribe a regular pentagon: divide the radius CD medially (II.59)in the point D F, so that CF may be the greater segment. Draw the radius CA at right angles to CD, and join A F Then, because the square of AF is greater than the square of CF by the square of the radius AC, and that CF is the side of the inscribed decagon (28.), AF is the side of the inscribed pentagon. Therefore, a chord, equal to A F, will subtend a fifth part of the circumference, and if the circumference be divided into five parts with chords each equal to AF, a regular pentagon will be inscribed, as required. 4. To inscribe a regular hexagon: divide the circumference into six parts with chords each equal to the radius (28). 5. To inscribe a regular decagon: divide the radius medially, and divide the circumference into ten parts with chords each equal to the greater segment of the radius SO divided (28). 6. To inscribe a regular pentedecagon : take the chord AB equal (as above) to the side of a regular inscribed pentagon, and the chord A D B E D equal (as above) to the side of an inscribed equilateral triangle: bisect the arc BD in E (54.), and join BE. Then, because the arc A B is contained in the whole circumference five times, and AD three times, if the circumference be divided into 5 x 3, or 15 equal parts, A B will contain three and AD five of those parts. Therefore the difference BD contains two of the same parts, and its half B E is one-fifteenth of the whole circumference. Divide the circumference, therefore, into fifteen parts with chords each equal to BE, and a regular pentedecagon will be inscribed, as required. And in every case, if through the angular points of the inscribed figure, or through the bisections of the arcs, (which is sometimes more convenient) there be drawn tangents to the circle (56.), a similar figure will be circumscribed about the circle (27.). Cor. Hence by the aid of II. 65. any one of the above-mentioned figures may be described upon a given finite straight line. Scholium. Besides the figures mentioned in the proposition, it has been discovered that any regular figure which has the number of its sides denoted by 2"+1 and prime, may be inscribed in a circle without any other aid than that of Plane Geometry, that is, by the intersections of the straight line and circle only. And it is evident that by dividing the sub |