PROBLEM B. If the given point is on the circumference, then the straight line drawn I to the radius through that point will be a tangent. If the given point is without the circle, join H the given point with the centre of the given circle : bisect CH in Q. With e as centre and radius QC or QH describe a circle cutting the given circle in T. Join HT; PROBLEM C. Between two points in the circumference of a circle find the shortest path which does not cut the circle. Let A, B be two points in the circumference of a 0. Then the arc AB which is the smaller of the two segments into which A, B divides the circumference shall be the shortest path between A and B which does not cut the o. For, if not, if possible, let there be a path, between A and B, passing through a point P outside the o which is as short as or shorter than any other : and let Q, R be the points where the path through P, on opposite sides of it, meets the Oce. Draw a tangent to the o at some point S in the circumference between Q and R cutting the path in T and V. Then TV is < the portion of the path TPV; .. a shorter path than that through P has been found; which is impossible, according to the hypothesis. Hence it follows that the arc AB is the shortest path between A and B which does not cut the circle. CONTACT OF CIRCLES. PROPOSITION VII. Circles which meet one another cannot have the same centre. For, if possible, let the circles AP, AQ meeting in A have a common centre C. Join AC, and draw any straight line CPQ cutting both circles. Then ::: C is the centre of O AP; .-. CA = CP; .. CP- CQ, which is impossible. PROPOSITION VIII. For, if possible, let the Os ABP, ABQ meet one another in A, B, C. Find O the centre of the OABP: join OA, OB, OC. Then OA, OB, OC are = one another. And :: from 0 three equal straight lines 0A, OB, OC are drawn to the circumference of O ABQ, ..O is the centre of O ABQ. (111. 3) ..O is the centre of two circles which cut one another, but this is impossible. (111. 7) .. One circle, &c. DEFINITION Two circles are said to touch one another when they are in contact, and one is entirely within or entirely without the other. PROPOSITION IX. One circle cannot touch another in more points than one. First, if possible, let one o ATBD touch another O AWBF internally in the points A, B. . Join AB; then the centres I, E of the interior and exterior circles respectively are in the straight line bisecting AB at right s. (III. 2) Produce EI to meet the circles in T and W. Join EB, IB. Then EI, IB are together > EB; (1. 16) .. EI, IT are together > EW; which is impossible. Next, if possible, let one O ABD touch another O ABF externally in the points A, B. Then the straight line joining A, B must be within each of the circles; (111. 1) which is impossible, since each falls entirely without the other. Hence, one circle cannot touch another in more points than one. Q.E.D. |