And since R, the point in which the circles nieet, is on the line of centres OD, .. the O PQR touches the given circle. Q.E.F. NOTE. There will be two solutions of this problem, since two circles may be drawn to touch EF, GH and to pass through D. a 25. To describe a circle to pass through a given point and touch a given straight line and a given circle. Through C draw DCEF perp. to A F and by describing a circle through F, E, and P, find a point K in DP (or DP produced) such that the rect. DE, DF =the rect. DK, DP. Describe a circle to pass through P, K, and touch AB: Ex. 21, p. 253. Join HE. III. 31. also the angle at F is a rt, angle; Constr. .. the points E, F, G, H are concyclic: .. the rect. DE, DF=the rect. DH, DG: III. 36. but the rect. DE, DF=the rect. DK, DP: Constr. the rect. DH, DG=the rect. DK, DP: the point H is on the O PKG. Let O be the centre of the O PHG. Join OG, OH, CH. and DG meets them. But since OG=OH, and CD=CH, . the LOGH=the L OHG; and the L CDH=the L CHD: the L OHG=the L CHD; OH and CH are in one st. line. :: the O PHG touches the given O DHE. Q. E. F. .. 1. 29. Notes. (i) Since two circles may be drawn to pass through P, K and to touch AB, it follows that there will be two solutions of the present problem. (ii) Two more solutions may be obtained by joining PE, and proceeding as before. The student should examine the nature of the contact between the circles in each case. a 26. Describe a circle to pass through a given point, to touch a given straight line, and to have its centre on another given straight line. 27. Describe a circle to pass through a given point, to touch a given circle, and to have its centre on a given straight line. 28. Describe a circle to pass through two given points, and to intercept an arc of given length on a given circle. 29. Describe a circle to touch a given circle and a given straight line at a given point. 30. Describe a circle to touch two given circles and a given straight line. We gather from the Theory of Loci that the position of an angle, line or figure is capable under suitable conditions of gradual change ; and it is usually found that change of position involves a corresponding and gradual change of magnitude. Under these circumstances we may be required to note if any situations exist at which the magnitude in question, after increasing, begins to decrease ; or after decreasing, to increase : in such situations the magnitude is said to have reached a Maximum or a Minimum value; for in the former case it is greater, and in the latter case less than in adjacent situations on either side. In the geometry of the circle and straight line we only meet with such cases of continuous change as admit of one transition from an increasing to a decreasing state-or vice versa—so that in all the problems with which we have to deal (where a single circle is involved) there can be only one Maximum and one Minimum—the Maximum being the greatest, and the Minimum being the least value that the variable magnitude is capable of taking. Thus a variable geometrical magnitude reaches its maximum or minimum value at a turning point, towards which the magnitude may mount or descend from either side : it is natural therefore to expect a maximum or minimum value to occur when, in the course of its change, the magnitude assumes a symmetrical form or position ; and this is usually found to be the case. This general connection between a symmetrical form or position and a maximum or minimum value is not exact enough to constitute a proof in any particular problem ; but by means of it a situation is suggested, which on further examination may be shewn to give the maximum or minimum value sought for. For example, suppose it is required to determine the greatest straight line that may be drawn perpendicular to the chord of a segment of a circle and intercepted between the chord and the arc : we immediately anticipate that the greatest perpendicular is that which occupies a symmetrical position in the figure, namely the perpendicular which passes through the middle point of the chord ; and on further examination this may be proved to be the case by means of 1. 19, and 1. 34. Again we are able to find at what point a geometrical magnitude, varying under certain conditions, assumes its Maximum or Minimum value, if we can discover a construction for drawing the magnitude so that it may have an assigned value : for we may then examine between what limits the assigned value must lie in order that the construction may be possible ; and the higher or lower limit will give the Maximum or Minimum sought for. It was pointed out in the chapter on the Intersection of Loci, (see page 125] that if under certain conditions existing among the data, two solutions of a problem are possible, and under other conditions, no solution exists, there will be some intermediate condition under which one and only one distinct solution is possible. Under these circumstances this single or limiting solution will always be found to correspond to the maximum or minimum value of the magnitude to be constructed. 1. For example, suppose it is required to divide a given straight line so that the rectangle contained by the two segments may be a maximum. We may first attempt to divide the given straight line so that the rectangle contained by its segments may have a given area—that is, be equal to the square on a given straight line. Let AB be the given straight line, and K the side of the given square. Y' z PZ It is required to divide the st. line AB at a point M, so that the rect. AM, MB may be equal to the sq. on K. Adopting a construction suggested by II. 14, describe a semicircle on AB; and at any point X in AB, or AB produced, draw XY perp. to AB, and equal to K. Through Y draw YZ par to AB, to meet the arc of the semicircle at P. Then if the perp. PM is drawn to AB, it may be shewn after the manner of 11. 14, or by III. 35 that the rect. AM, MB=the sq. on PM = the sq. on K. So that the rectangle AM, MB increases as K increases. Now if K is less than the radius CD, then YZ will meet the arc of the semicircle in two points P, P'; and it follows that AB may be divided at two points, so that the rectangle contained by its segments may be equal to the square on K. If K increases, the st. line YZ will recede from AB, and the points of intersection P, P' will con. tinually approach one another; until, when K is equal to the radius CD, the st. line YZ (now in the position Y'Z') will meet the arc in two coincident points, that is, will touch the semicircle at D; and there will be only one solution of the problem. If K is greater than CD, the straight line YZ will not meet the semicircle, and the problem is impossible. Hence the greatest length that k may have, in order that the construction may be possible, is the radius CD. :: the rect. AM, MB is a maximum, when it is equal to the square on CD; that is, when PM coincides with CD, and consequently when M is the middle point of AB. NOTE. The special feature to be noticed in this problem is that the maximum is found at the transitional point between two solutions and no solution ; that is, when the two solutions coincide and be. conie identical. a The following example illustrates the same point. 2. To find at what point in a given straight line the angle subtended by the line joining two given points, which are on the same side of the given straight line, is a maximum. Let CD be the given st. line, and A, B the given points on the same side of CD. It is required to find ut what point in CD the angle subtended by the st. line AB is a maximum. First determine at what point in CD, the st. line AB subtends a given angle. This is done as follows:- On AB describe a segment of a circle containing an angle equal to the given angle. III. 33. If the arc of this segment intersects CD, two points in CD are found at which AB subtends the given angle : but if the arc does not meet CD, no solution is given. In accordance with the principles explained above, we expect that a maximum angle is determined at the limiting position ; that is, when the arc touches CD, or meets it at two coincident points. [See page 231.] This we may prove to be the case. Describe a circle to pass through A and B, and to touch the st. line CD. [Ex. 21, p. 253.] Let P be the point of contact. Then shall the 4 APB be greater than any other angle subtended by AB at a point in CD on the same side of AB as P. For take Q, any other point in CD, on the same side of AB as P; and join AQ, QB. B Since Q is a point in the tangent other than the point of contact, it must be without the circle ; .. either BQ or AQ must meet the arc of the segment APB. Let BQ meet the arc at K: join AK. Then the L APB=the L AKB, in the same segment: but the ext. L AKB is greater than the int. opp. L AQB. : the L APB is greater than AQB. Similarly the L APB may be shewn to be greater than any other angle subtended by AB at a point in CD on the same side of AB: that is, the 2 APB is the greatest of all such angles. Q.E.D. NOTE. Two circles may be described to pass through A and B, and to touch CD, the points of contact being on opposite sides of AB; |