Theorem 19. If two circumferences meet in a point which is not on their center-line, then (1) they meet in one other point, (2) their center-line is the perpendicular bisector of their common chord, (3) their centersegment is greater than the difference and less than the sum of the radii. M M N Given M and N, two circumferences with centers A, B, meeting at P not on AB. To prove that (1) they meet again, as at P', (2) AB ¦ PP' and bisects it, as at C, (3) AB > the difference between AP and BP and <AP+ BP. Proof. 1. In Fig. 1, suppose ▲ ABP revolved about AB as an axis of symmetry, thus determining ▲ AP'B. 2. Then... AP' = AP, and BP'= BP, 3. ... P' is on both M and N, which proves (1). Step 1 Def. O, cor. 4 4. In Fig. 2, .A ABPA ABP', Step 1 AC. 6... A ACPA ACP', and PP' is bisected at rt. by AB, which proves (2). 7. AB the difference between <AP + BP, which proves (3). I, th. 1 AP and BP and I, th. 8 and cor. COROLLARY. If two circumferences meet at one point only, that point is on their center-line. (Why ?) EXERCISE. 326. If two circumferences intersect, any two parallel lines drawn through the points of intersection and terminated by the respective circumferences are equal. Theorem 20. If two circles meet on their center-line, they are tangent. P Given O and O', the centers of two circles with radii OA, O'A, which meet on their center-line at A. To prove that the circles are tangent. Proof. 1. Let P be any point, other than A, on circumference with center O, and draw OP, O'P. 2. Then I, th. 8 OA - O'A. OO' + O'P > OP or its equal OA. 4. COROLLARIES. Axs. 4, 5 Def. O, cor. 4 they have only one point Def. 1. If two circumferences intersect, neither point of intersection is on the center-line. (Why ?) 2. If two circles touch, they have a common tangent-line at the point of contact. For a perpendicular to their center-line at that point is tangent to both. (Why?) EXERCISES. 327. If the center-segment of two circles is (1) greater than, (2) equal to, (3) less than the sum of the two radii, the circumferences (1) do not meet, (2) are tangent, (3) intersect. 328. The greatest of all lines joining two points, one on each of two given circumferences, is greater than the center-segment by the sum of the radii. 329. If two circles, whose centers are O, O', are tangent at P, and a line through P cuts the circumferences at A, A', prove that OA | O'A'. Two cases; external and internal tangency. Show that the proposition is true for any number of circles. 330. Find the locus of the centers of all circles tangent to a given circle at a given point. Section 6. Problems. Problem 1. To bisect a given arc. Solution. 1. Draw its chord AB. Post. 2 I, pr. 4 A 2. Draw PC LAB at its mid-point. 3. Then PC bisects the arc. Th. 5, cor. 2 EXERCISES. 331. To trisect a quadrant. (See suggestion on p. 67, ex. 194.) 332. To double a given arc of a given circumference. Suppose the arc is equal to or greater than a semicircumference. 333. Through a point in a circle to draw a chord which shall be bisected by that point. Problem 2. To find the center of a circle, given its circumference or any arc. Given a circumference, or an arc ABC. 2. Draw their perpendicular bisec Proof. DD' and EE' intersect at the center O. I, th. 31, cor. 1 NOTE. Hereafter it will be assumed that the center is known if an arc is known, for it may always be found by this problem. EXERCISES. 334. To find the locus of the mid-points of a system of parallel chords of a circle. 335. Given two arcs of circles, show how to find if they are arcs of (1) the same circle, (2) concentric circles. Problem 3. To draw a tangent to a given circle from a given point. 1. If the point is on the circumference. Solution. 1. At the given point erect a perpendicular to the radius drawn to the point. I, pr. 2 2. This is the required tangent, and the solution is 3. Describe a O with center M, radius MO. Post. 6 4. Join A to intersections of circumferences. Post. 2 5. Then these lines, AP, AP', are the required tangents. Proof. 1. The circumferences will have two points in common, DEFINITIONS. Two intersecting arcs are said to form an angle, meaning thereby the angle included by their respective tangents at the point of intersection. An arc and a secant are said to form an angle, meaning thereby the angle included by the secant and the tangent to the arc at the point of meeting. E.g. in the figure of pr. 4, OP is said to make a right angle with the circumference PP'B, because it is perpendicular to the tangent at P. EXERCISE. 336. If step 4 of the construction of pr. 3 read, "From A draw a perpendicular to OP," show that the solution would be valid. Show also that this solution would answer for both cases (1) and (2), since (1) is then only a special case of (2), A having moved to the circumference. |
