Abbildungen der Seite
PDF
EPUB

33. Draw two radii containing an angle equal to the supplement of the given angle; the tangents drawn at the extremities of these radii will contain the given angle.

34. Since the circle is to touch two parallel lines drawn from two given points in a third line, the radius of the circle is determined by the distance between the two given points.

35. It is sufficient to suggest that the angle between a chord and a tangent is equal to the angle in the alternate segment of the circle. Euc. III. 32.

36. Let AB be the given chord of the circle whose center is O. Draw DE touching the circle at any point E and equal to the given line; join DO, and with center O and radius DO describe a circle: produce the chord AB to meet the circumference of this circle in F: then F is the point required.

37. Let D be the point required in the diameter BA produced, such that the tangent DP is half of DB. Join CP, C being the center. Then CPD is a right-angled triangle, having the sum of the base PC and hypotenuse CD double of the perpendicular PD.

38. If BE intersect DF in K (fig. Euc. III. 37). Join FB, FE, then by means of the triangles, BE is shewn to be bisected in K at right angles.

39. Let AB, CD be any two diameters of a circle, O the center, and let the tangents at their extremities form the quadrilateral figure EFGH. Join EO, OF, then EO and OF may be proved to be in the same straight line, and similarly HO, OK.

NOTE.-This Proposition is equally true if AB, CD be any two chords whatever. It then becomes equivalent to the following proposition:The diagonals of the circumscribed and inscribed quadrilaterals, intersect in the same point, the points of contact of the former being the angles of the latter figure.

40. Let C be the point without the circle from which the tangents CA, CB are drawn, and let DE be any diameter, also let AE, BĎ be joined, intersecting in P, then if CP be joined and produced to meet DE in G: CG is perpendicular to DE. Join DA, EB, and produce them to

meet in F.

Then the angles DAE, EBD being angles in a semicircle, are right angles; or DB, EA are drawn perpendicular to the sides of the triangle DEF: whence the line drawn from F through P is perpendicular to the third side DE.

41. Let the chord AB, of which P is its middle point, be produced both ways to C, D, so that AC is equal to BD. From C, D, draw the tangents to the circle forming the tangential quadrilateral CKDR, the points of contact of the sides, being E, H, F, G. Let O'be the center of the circle. Join EH, GF, CO, GO, FO, DO. Then EH and GF may be proved each parallel to CD, they are therefore parallel to one another. Whence is proved that both EF and DG bisect AB.

42. This is obvious from Euc. 1. 29, and the note to г. 22. p. 156. 43. From any point A in the circumference, let any chord AB and tangent AC be drawn. Bisect the are AB in D, and from D draw DE, DC perpendiculars on the chord AB and tangent AC. Join AD, the triangles ADE, ADC may be shewn to be equal.

44. Let A, B, be the given points. Join AB, and upon it describe a segment of a circle which shall contain an angle equal to the given angle. If the circle cut the given line, there will be two points; if it only touch the line, there will be one; and if it neither cut nor touch the line, the problem is impossible.

45. It may be shewn that the point required is determined by a perpendicular drawn from the center of the circle on the given line.

46. Let two lines AP, BP be drawn from the given points A, B, making equal angles with the tangent to the circle at the point of contact P, take any other point Q in the convex circumference, and join QA, QB: then by Prob. 4, p. 71, and Euc. 1. 21.

47. Let C be the center of the circle, and E the point of contact of DF with the circle. Join DC, CE, CF.

48. Let the tangents at E, F meet in a point R. Produce RE, RF to meet the diameter AB produced in S, T. Then RST is a triangle, and the quadrilateral RFOE may be circumscribed by a circle, and RPO may be proved to be one of the diagonals.

49. Let C be the middle point of the chord of contact: produce AC, BC to meet the circumference in B', A', and join AA', BB'.

50. Let A be the given point, and B the given point in the given line CD. At B draw BE at right angles to CD, join AB and bisect it in F, and from F draw FE perpendicular to AB and meeting BE in E. E is the center of the required circle.

51. Let O be the center of the given circle. Draw OA perpendicular to the given straight line; at O in OA make the angle AOP equal to the given angle, produce PO to meet the circumference again in Q. Then P, Qare two points from which tangents may be drawn fulfilling the required condition.

52. Let C be the center of the given circle, B the given point in the circumference, and A the other given point through which the required circle is to be made to pass. Join CB, the center of the circle is a point in CB produced. The center itself may be found in three ways.

53. Euc. III. 11 suggests the construction.

54. Let AB, AC be the two given lines which meet at A, and let D be the given point. Bisect the angle BAC by AE, the center of the circle is in AE. Through D draw DF perpendicular to AE, and produce DF to G, making FG equal to FD. Then DG is a chord of the circle, and the circle which passes through D and touches AB, will also pass through G and touch AC.

55. As the center is given, the line joining this point and center of the given circle, is perpendicular to that diameter, through the extremities of which the required circle is to pass.

56. Let AB be the given line and D the given point in it, through which the circle is required to pass, and AC the line which the circle is to touch. From D draw DE perpendicular to AB and meeting AC in C. Suppose O a point in AD to be the centre of the required circle. Draw OE perpendicular to AC, and join OC, then it may be shown that CO bisects the angle ACD.

57. Let the given circle be described. Draw a line through the center and intersection of the two lines. Next draw a chord perpendicular to this line, cutting off a segment containing the given angle. The circle described passing through one extremity of the chord and touching one of the straight lines, shall also pass through the other extremity of the chord and touch the other line.

58. The line drawn through the point of intersection of the two circles parallel to the line which joins their centers, may be shewn to be double of the line which joins their centers, and greater than any other straight line drawn through the same point and terminated by the circumferences. The greatest line therefore depends on the distance between the centers of the two circles.

59. Apply Euc. III. 27. 1. 6.

60. Let two unequal circles cut one another, and let the line ABC drawn through B, one of the points of intersection, be the line required, such that AB is equal to BC. Join O, O' the centers of the circles, and draw OP, O'P' perpendiculars on ABC, then PB is equal to BP'; through O' draw O'D parallel to PP'; then ODO' is a right-angled triangle, and a semicircle described on OO' as a diameter will pass through the point D. Hence the synthesis. If the line ABC be supposed to move round the point B and its extremities A, C to be in the extremities of the two circles, it is manifest that ABC admits of a maximum.

61. Suppose the thing done, then it will appear that the line joining the points of intersection of the two circles is bisected at right angles by the line joining the centers of the circles. Since the radii are known, the centers of the two circles may be determined.

Join

62. Let the circles intersect in A, B; and let CAD, EBF be any parallels passing through A, B and intercepted by the circles. CE, AB, DF. Then the figure CEFD may be proved to be a parallelogram. Whence CAD is equal to EBF.

63. Complete the circle whose segment is ADB; AHB being the other part. Then since the angle ACB is constant, being in a given segment, the sum of the arcs DE and AHB is constant. But AHB is given, hence ED is also given and therefore constant.

64. From A suppose ACD drawn, so that when BD, BC are joined, AD and DB shall together be double of AC and CB together. Then the angles ACD, ADB are supplementary, and hence the angles BCD, BDC are equal, and the triangle BCD is isosceles. Also the angles BCD, BDC are given, hence the triangle BDC is given in species. Again AD + DB

=

2.AC 2. BC, or CD

=

AC + BC.

с

Whence, make the triangle bde having its angles at d, c equal to that in the segment BDA; and make ca = cd cb, and join ab. At A make

the angle BAD equal to bad, and AD is the line required.

65. The line drawn from the point of intersection of the two lines to the center of the given circle may be shewn to be constant, and the center of the given circle is a fixed point.

66.

This is at once obvious from Euc. III. 36.

67. This follows directly from Euc. III. 36.

68. Each of the lines CE, DF may be proved parallel to the common chord AB.

69. By constructing the figure and joining AC and AD, by Euc. III. 27, it may be proved that the line BC falls on BD.

70. By constructing the figure and applying Euc 1. 8, 4, the truth is manifest.

71. The bisecting line is a common chord to the two circles; join the other extremities of the chord and the diameter in each circle, and the angles in the two segments may be proved to be equal,

72. Apply Euc. III. 27; 1. 32, 6.

73. Draw a common tangent at C the point of contact of the circles, and prove AC and CB to be in the same straight line.

74. Let A, B, be the centers, and C the point of contact of the two circles; D, E the points of contact of the circles with the common tangent DE, and CF a tangent common to the two circles at C, meeting DF in E. Join DC, CE. Then DF, FC, FE may be shewn to be equal, and FC to be at right angles to AB.

75. The line must be drawn to the extremities of the diameters which are on opposite sides of the line joining the centers.

76. The sum of the distances of the center of the third circle from the centers of the two given circles, is equal to the sum of the radii of the given circles, which is constant.

77. Let the circles touch at C either externally or internally, and their diameters AC, BC through the point of contact will either coincide or be in the same straight line. CDE any line through C will cut off similar segments from the two circles. For joining AD, BE, the angles in the segments DAC, EBC are proved to be equal.

The remaining segments are also similar, since they contain angles which are supplementary to the angles DAC, EBC.

78. Let the line which joins the centers of the two circles be produced to meet the circumferences, and let the extremities of this line and any other line from the point of contact be joined. From the center of the larger circle draw perpendiculars on the sides of the right-angled triangle inscribed within it.

79. In general, the locus of a point in the circumference of a circle which rolls within the circumference of another, is a curve called the Hypocycloid; but to this there is one exception, in which the radius of one of the circles is double that of the other: in this case, the locus is a straight line, as may be easily shewn from the figure.

80. Let A, B be the centers of the circles. Draw AB cutting the circumferences in C, D. On AB take CE, DF each equal to the radius of the required circle: the two circles described with centers A, B, and radii AE, BF, respectively, will cut one another, and the point of intersection will be the center of the required circle.

81. Apply Euc. III. 31.

82. Apply Euc. III. 21.

83. (1) When the tangent is on the same side of the two circles. Join C, C their centers, and on CC' describe a semicircle. With center C' and radius equal to the difference of the radii of the two circles, describe another circle cutting the semicircle in D: join DC and produce it to meet the circumference of the given circle in B. Through C draw CA parallel to DB and join BA; this line touches the two circles.

(2) When the tangent is on the alternate sides. Having joined C, C'; on CC' describe a semicircle; with center C, and radius equal to the sum of the radii of the two circles describe another circle cutting the semicircle in D, join CD cutting the circumference in A, through C draw CB parallel to CA and join AB.

84. The possibility is obvious. The point of bisection of the segment intercepted between the convex circumferences will be the center of one of the circles: and the center of a second circle will be found to be the point of intersection of two circles described from the centers of the given circles with their radii increased by the radius of the second circle. The line passing through the centers of these two circles will be the locus of the centers of all the circles which touch the two given circles.

85. At any points P, R in the circumferences of the circles, whose centers are A, B, draw PQ, RS, tangents equal to the given lines, and join AQ, BS. These being made the sides of a triangle of which AB is the base, the vertex of the triangle is the point required.

86. In each circle draw a chord of the given length, describe circles concentric with the given circles touching these chords, and then draw a straight line touching these circles.

87. Within one of the circles draw a chord cutting off a segment equal to the given segment, and describe a concentric circle touching the chord: then draw a straight line touching this latter circle and the other given circle.

88. The tangent may intersect the line joining the centers, or the line produced. Prove that the angle in the segment of one circle is equal to the angle in the corresponding segment of the other circle.

89. Join the centers A, B; at C the point of contact draw a tangent, and at A draw AF cutting the tangent in F, and making with CF an angle equal to one-fourth of the given angle. From F draw tangents to the circles.

90. Let C be the center of the given circle, and D the given point in the given line AB. At D draw any line DE at right angles to AB, then the center of the circle required is in the line AE. Through C draw a diameter FG parallel to DE, the circle described passing through the points E, F, G will be the circle required.

91. Apply Euc. f. 18.

92. Let A, B, be the two given points, and C the center of the given circle. Join AC, and at C draw the diameter DCE perpendicular to AC, and through the points A, D, E describe a circle, and produce AC to meet the circumference in F. Bisect AF in G, and AB in H, and draw GK, HK, perpendiculars to AF, AB respectively and intersecting in K. Then K is the center of the circle which passes through the points A, B, and bisects the circumference of the circle whose center is C.

93. Let D be the given point and EF the given straight line. (fig. Euc. 11. 32.) Draw DB to make the angle DBF equal to that contained in the alternate segment. Draw BA at right angles to EF, and DA at right angles to DB and meeting BA in A. Then AB is the diameter of the circle.

94. Let A, B be the given points, and CD the given line. From E the middle of the line AB, draw EM perpendicular to AB, meeting CD in M, and draw MA. In EM take any point F; draw FH to make the given angle with CD; and draw FG equal to FH, and meeting MA produced in G. Through A draw AP parallel to FG, and CPK parallel to FH. Then P is the center, and C the third defining point of the circle required: and AP may be proved equal to CP by means of the triangles GMF, AMP; and HMF, CMP, Euc. vI. 2. Also CPK the diameter makes with CD the angle KCD equal to FHD, that is, to the given angle.

95. Let A, B be the two given points, join AB and bisect AB in C, and draw CD perpendicular to AB, then the center of the required circle will be in CD. From O the center of the given circle draw CFG parallel to CD, and meeting the circle in F and AB produced in G. At F draw a chord FF equal to the given chord. Then the circle which passes through the points at B and F, passes also through F'.

96. Let the straight line joining the centers of the two circles be produced both ways to meet the circumference of the exterior circle.

97. Let A be the common center of two circles, and BCDE the chord such that BE is double of CD. From A, B draw AF, BG perpendicular to BE. Join AC, and produce it to meet BG in G. Then AC may be shewn to be equal to CG, and the angle CBG being a right angle, is the angle in the semicircle described on CG as its diameter.

98. The lines joining the common center and the extremities of the chords of the circles, may be shewn to contain unequal angles, and the angles at the centers of the circles are double the angles at the circumferences, it follows that the segments containing these unequal angles are not similar.

99. Let AB, AC be the straight lines drawn from A, a point in

« ZurückWeiter »