B I N B touch one, two, or three given circles, after a similar manner; viz. by describor which shall touch both a straight ing a circle which shall touch the given line and a circle, or two straight lines straight line in the given point, and likeand a circle, or a straight line and two wise pass through a point assumed in circles, and at the same time pass the circumference of the circle, and then through one or two given points, as proceeding as in the proposition. the other data may happen to admit. 2. To describe a circle which shall The six new problems of contact which pass through a given point A, and are thus suggested are too remarkable touch two given circles B C D E F G. to be passed over without further no Suppose that the required circle is tice; they are accordingly here sub- described, and that it touches the given joined. circles in the points D, F respectively: 1. To describe a circle which shall join D F, and since the straight line pass through two given points A, B, D F cannot touch the given circles in and touch a given circle CDE. the points D, F, (because then it would 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 H M meet the circumference CDE in no other point, it is the circle re .A quired : but if it do, let that other touch the required circle ADF in the point be D: join same points, which (1. Cor. 2.) is absurd) A B, CD, and let it be produced both ways to meet the let them be pro circumferences a second time in the duced to meet in points C, G respectively: take H, K, the F: from F draw centres of the circles BCD, EFG, and FG, touching the join HK, HD, KF,KG: then, because circle CDE (56.): the circle ADF touches BCD in D, let G be the point and EFG in F, the radii HD, KF of contact, and produced (8. Cor. 1.) will meet in its describe a circle (59.) passing through centre, L; and, because KGF, LFD the three points A, B, G. Then, be are isosceles triangles, the angle KGF cause the chord C D of the circle is (I. 6.) equal to KFG, that is (I. 3.) CDE meets the tangent GF in F to L FD, that is again (I. 6.) to LDF (21.), the square of GF is equal to the or (I. 3.) HDC: therefore (I. 15.) KG rectangle CF, FD, that] is, to the is parallel to HD, and, consequently, if rectangle A F, FB (20.): therefore the circles B C D, EF G be equal, (21. Cor.) GF touches also the circle H K, D G will be parallel (I. 21.), or if ABG, and, consequently, the circle one of them, as BCD, be greater ABG touches the circle ČDE (2. Cor. than the other, H K and D G will meet, 2 and 9.), and is the circle required. if produced in some point M. In the If A B and C D be parallels (which latter case, draw MB touching the will be the case when the line which bi- circle BCD in B (56.), join H B, and sects A B at right angles passes through from K draw KI (I. 45.) perpendicular the centre of the circle CDE), a tan- to MB, and therefore (I. 14.) parallel gent FG is to be drawn parallel to to HB: then, because KI, H B are A B or CD (57.), and the circle ABG, parallel, KI : HB::KM : HM (II.30. being then described as before, will be Cor. 2.) that is, :: KG:HD, because the circle required. KG, HD are parallel: but HB is It is evident, in each case, that there equal to HD; therefore (II. 18.) K I is are two tangents FG, and two circles equal to KG, and I is a point in the cirABG corresponding to them, one cumference of the circle E FG, and (2.) touching the given circle externally, MB touches the circle E F G in the the other internally. point I. Now, MI: MB :: MK: Cor. The problem which requires a MH (II. 29.), that is, :: MG: MD; circle to be described which shaīl touch therefore, alternando (II. 19.), MI : MG a given straight line in a given point, :: MB : MD: but, because MI touches and also a given circle, may be solved the circle EFG, the square of MI is F B С equal to the rectangle M Fx MG (21.), quired, to which the reader will be and consequently (11.38. Cor. 1.) MF: easily conducted by an investigation MI:: MI : MG: therefore (II. 12.) similar to that which has been already MF:MI:: MB:MD, and (38.) the given. rectangle MIX M B is equal to the rect 3. To describe a circle which shall angle M Fx M D. Join M A, and let touch three given circles A, B, C. it cut the circumference of the circle Let C be that one of the three which A FD a second time in the point is not greater than either of the other N: then the rectangle M A M N is two. From the centres of A and B equal (20.) to the rectangle MFx describe two new circles with radii less MD, that is, by what has been just than their own respectively by the rademonstrated, to the rectangle MIX A 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 H K which joins their centres in M: join MA: take M N a fourth proportional (II. 53.) to MA, MB, M I, and, as in (1), describe the circle AND, passing through the two points A, N and dius of C; and, as in the last problem, touching the circle BCD: AND will describe a circle D E F which shall pass be the circle required. For if MD through the centre F of the circlc C, and be joined, cutting the circle EFG in touch these two new circles in the points F and G, and if K F and KG be joined, D, E respectively: then, if a circle be dethe circle described will (20. Cor.) pass scribed concentric with the circle DEF, through F, because (as has been shown with a radius less than that of DEF above) MDX M F is equal to M Bx by the radius of C, it will evidently touch MI, that is to MAXMN; and (which each of the three given circles (10. Cor. has been likewise shown) K G will be pa- 2.), and will be the circle required. rallel to HD. Therefore, if in H D pro In the case here considered, the reduced (which(8. Cor. 1.) passes through quired circle touches each of the given the centre of the circle AND, because circles externally: since, however, a circle it passes through the centre H of B CD, may be described which shall touch all which touches AND in D), there be three internally, or any two externally, taken L, the centre of the circle AND, and the third internally, or again, any and L F be joined, the angle LFD will two internally and the third externally, be equal (I. 6.) to LDF, that is (I. 3.) there are no less than eight different cirto HD C, that is, again (I. 15.) to KGF, cles which satisfy the problem: the or (I. 6.) KFG; and because DFG is construction of each of these is oba straight line, K F and F L likewise lie tained after a similar manner. in a straight line (1. 3.), and therefore 4. To describe a circle which shall the circle AND touches the ircle EFG pass through a given point A, touch a in F (9.). given straight line B C, and also touch Should the given circles be equal, or a given circle D E F. should the point N coincide with the Take G the centre (55.) of the circle point A, the foregoing solution must be DEF, and through G draw (I. 45.) modified accordingly. The latter case the diameter D F perpendicular to BC is provided for by the corollary of the to meet B C in C: join A D, and take preceding problem: both, indeed, are DH a fourth proportional (II. 53.) to comparatively easy, and are therefore AD, DC and D F, i. e. (II. 38.) such left to the student. that A DxDH may be equal to D Cx 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 DF; and describe a circle H A B (59.) these a clifferent construction is re pussing throu h the two points A,, D E L F в and touching the straight line BC: (55.): and (59.) describe the circle this shall be the circle required. For, HLN passing through the point N let it touch B C in B; join BD, and and touching the straight lines HK, let it cut the circumference DEF in LM: then, if a circle be described K, and join KF: then, because (15. concentric with H LN, and with a raCor. 1.) Ď KF is a right angle, the tri dius less than that of HLN by the angles D KF, D C B are equiangular, radius of EFG, it will evidently touch and (II. 31.) DK:DF::DC:D B, or the straight lines AB, and CD, and (38.) D KXD B is equal to D CRD F, the circle EFG (10. Cor. 2.) and will that is, to ADxDH; therefore K is therefore be the circle required. also a point in the circumference of the Four different circles may be described circle À BH (20. Cor.). Now take L to satisfy this problem, viz. two touchthe centre of the circle A BH (55.) and ing the given circle externally, and two join LB, LK, GK: and, because LB internally:* the latter two may be found is (2. Cor. 1.) perpendicular to BC, it by drawing H K and L M between A B is (I. 14.) parallel to D C, and therefore and C D. (I. 15.) the angle LBK is equal to 6. To describe a circle which shall GDK; but LK B is equal to LBK, touch a given straight line A B, and and GKD to GDK, because L B is two given circles CDE, FGH. equal to L K, and G D to G K (I. 6.); If the given circles be equal to one therefore the angle LKB is equal to another, draw K L parallel to A B, at GKD, and (I. 3.) LKG is a straight a distance from it equal to their comline: therefore the circle ABH touches mon radius (I. 48.), and (59.) describe (9.) the circle D E F in the point K, a circle passing through the centres of and is the circle required. the two given circles and touching the If DH be equal to DA, a circle line K L: then it is evident that a cirABH is to be described touching D A cle concentric with the latter, and havin A, and also touching the straight ing a radius less by the common radius line BC. (See the beginning of this of the given circles, will be the circle Scholium.) required. But, if one of them, as In this problem there are two circles FGH, be greater than the other, desatisfying the conditions : the construction of that which touches the given circle internally will readily be understood by applying what has been said H to the subjoined figure. A B K. scribe about its centre a cirele f g h with a radius less than that of FGH 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 5. To describe a circle which shall the centre of C DE touching the line touch two given straight lines A B, KL, and touching also the circle fgh, CD, and also a given circle E F G. as shown in the last problem but one: Draw HK parallel to A B, and at a then, if a circle be described concentric distance from it equal to the radius of with this and with a radius less by the EFG: draw also L M parallel to and radius of C D 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 K D L G K к H B Н. N to the construction of them all. * Erternal contact is where two circles touch one another, and each of them is without the other; inD ternal contact, where two circles touch one another, and one of them is within the other. In the latter M L case, it is evident that the circumference of the larger circle is without the circumference of the at the same distance from CD (I. 48.): other; and yet each of them is properly said to touch ke N the centre of the circle EFG tion of contact which is called internal. the other internally, meaning, with that descrip. B В D B Much as each of the foregoing pro- PROP. 62. Prob. 9. (Euc. iv. 2, 3.) blems may be varied in appearance by the consideration of internal contact, Given a circle A BC, to (1) inscribe they admit of being varied yet further in it, or (2) describe about it, a triangle by supposing, when a point is given, similar to a given triangle DEF. that it is in the given tangent or cir 1. Let A B C be the required incumference: and when it is besides sug. scribed triangle, and through A draw the gested that, in each of the ten problems line G AH touching the circle (56.): furnished by the proposition and scho- then the angle GAB is equal to the anlium, one of the data be changed for gle at C, and the angle H A C to the an"a given radius," or for“ a given line gle at B (17.) Therefore, reversely, take in which the centre is to lie," or again, any point A of the circumference, draw two of the data for both of these, the the tangent GAH (56.), and make the student will have before him a field no angles GAB, HAC equal to the angles 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 at F and E respectively. (I. 47.) Then, shall contain an angle equal to a given if A B, A C meet the circumference in angle ABC. Let A Q B be the segment required, be joined, A B C will be the triangle the points B, C respectively, and if BC and P its centre. Then, because the angle required; for, the angles at B and c A B C is equal to the angle in the al- being equal to the angles at E and F; ternate segment, the line B C touches each to each, the third angles at A and the circle (17. Cor.) at B. Therefore, PB D are likewise equal (1. 19. Cor. 1.), joined is at right angles to B C. Again, and therefore the triangles are similar because P is the centre (II. 31. Cor. 1.). of a circle having the chord 2. Let KLM be the required circumB, is in the straight line which JP scribed triangle, and let its sides touch bisects A B at right an the circle in the points A, B, C: take O the centre of the circle (55.), and join gles (3 Cor. 2). There- A fore, reversely, draw ОА, ОВ, ОС. BD perpendicular to BC (1.44.), and bisect A B at right angles A (1.43.Cor.) by a straight line, cutting BD in P: with the centre Pand radius PB describe the circular arc BQA; and the segment BQ A will be the segment required. Then, because the angles at A and B of the quadrilateral ALBO are right PROP. 61. Prob. 8. (Euc. iii. 34.) angles (2. Cor. 1.), the remaining angles Given a point A in the circumference L and A O B are together equal to two of a circle A BC; to cut off a segment right angles (I. 20. Cor.) and AOB is which shall contain a given angle, by the supplement of the angle L, or, of a straight line drawn from the point A. its equal, the angle E; which supple From A draw A D touching the cir- ment may be obtained by producing the cle, and let A B be the line required: side E F. In like manner it may be then, because the angle shown, that the angle B O C is the supBAD (17.) is equal plement of the angle M or F. Therefore, to the angle in the al reversely, taking any point A'in the cirternate segment BCA, cumference, at the point 0, make (1.47.) it is equal to the given the angle A O B equal to the supplement angle. Therefore, re of E, and the angle BOC equal to the versely, make DAB supplement of F: at the points A, B, C equal to the given an draw tangents (56.) meeting one another gle (I. 47); and AB in K, L, M ; and KLM will be the triwill be the line required. angle required. B DP M A B B D a regular pentagon will be inscribed, as PROP. 63. Prob. 10. (Euc. iv. 6, 7, required. 11, 12, 15, 16.) 4. To inscribe a reguA circle being given ; to inscribe lar hexagon: divide the in it, or describe about it, circumference into six 1. an equilateral triangle; or parts with chords each 2. a square ; or equal to the radius (28). 3. a regular pentagon; or 4. a regulur hexagon ; or 5. To inscribe a regular decagon: di5. a regular decagon; or vide the radius medially, and divide the 6. a regular pentedecagon. circumference into ten Take C the centre of the circle; and, parts with chords 1. to inscribe an equilateral triangle: each equal to the from A, any point of the circumference, greater segment of the with the radius A C, describe a circle radius divided cutting the given circle in the points (28). B, D: produce A C to meet the circumference 6. To inscribe a rein E, and join BD, gular pentedecagon: DE, E B. Then join take the chord AB ing BC, CD, BA, AD, equal (as above) to c because the triangles B the side of a regular ACB, ACD are equi inscribed pentagon, lateral, the angles and the chord A D BCA,'D CA are each of them a third equal (as above) to the side of an inof two right angles (I. 6. and I. 19.). scribed equilateral triangle: bisect the Therefore the adjacent angles B CE, arc BD in E (54.), and join BE. Then, DCE are each of them two-thirds of the because the arc A B is contained same, as is also the angle BCD; and in the whole circumference five times, because the sides of the triangle B D E and AD three times, if the circumfersubtend equal angles at the centre C, ence be divided into 5 x 3, or 15 equal they are equal to one another (12. Cor.1.), parts, A B will contain three and i.e. the triangle B D E is equilateral. AD five of those parts. Therefore 2. To inscribe a the difference B D contains two of the square : draw two dia same parts, and its half B E is one-fifmeters at right angles teenth of the whole circumference. Dito one another, and vide the circumference, therefore, into join their extremities: fifteen parts with chords each equal to the included figure will BE, and a regular pentedecagon will be a square; for its be inscribed, as required. sides are equal, because they subtend And in every case, if through the anequal angles at the centre C(12. Cor. 1.), gular points of the inscribed figure, and its angles are right angles, because or through the bisections of the arcs, they are contained in semicircles (15. (which is sometimes more convenient) Cor. 1.) there be drawn tangents to the circle 3. To inscribe a re (56.), a similar figure will be circumgular pentagon:divide scribed about the circle (27.). the radius CD medi Cor. Hence by the aid of II. 65. any ally(11.59)in the point D one of the above-mentioned figures F, so that CF may be may be described upon a given finite the greater segment. straight line. Draw the radius CA at right angles to CD, and join AF. Scholium. Then, because the square of_AF is Besides the figures mentioned in the greater than the square of C F by the proposition, it has been discovered that square of the radius AC, and that CF is any regular figure which has the numthe side of the inscribed decagon (28.), ber of its sides denoted by 2n + 1 and AF is the side of the inscribed pentagon. prime, may be inscribed in a circle withTherefore, a chord, equal to A F, will out any other aid than that of Plane subtend a fifth part of the circumference, Geometry, that is, by the intersections of and if the circumference be divided into the straight line and circle only. And five parts with chords each equal to AF, it is evident that by dividing the sub |