Solutions 2 MT1100 From Euclid to Mandelbrot 1. The equations of C1 and C2 are (x - 1)2 + (y - 2)2 = 9. x2 + y2 = 4 and (1) (2) Multiplying out (2) and simplifying, we find x2 - 2x + y2 - 4y = 4. Subtracting (1) we find that -2x - 4y = 0 and so x = - 2y. Substituting this value of x into (1) we find that 5y2 = 4 and so y = +2/5. Since x = - 2y, the points of intersection are therefore (-4/v5, 2/v5) and (4/15, -2/v/5). 2. The equations of the line and circle are y = ax and (x - 7)2 + (y + 1)2 = 25, so the line and circle intersect when (x - 7)2 + (ax + 1)2 = 25. Multiplying out and simplifying we get the quadratic equation (1+ a2)x2 + (2a - 14)x + 25 = 0. If there is only one point of intersection, we must have just one solution x to this quadratic equation, and so 'b2 - 4ac = 0' gives a condition that a must satisfy: (2a - 14)2 - 4(1 + a2) · 25 = 0, or -96a2 - 56a + 96 = 0. This quadratic equation has two solutions: a = - 4/3 and a = 3/4, and these are the values of a where our circle and line intersect just once. For a diagram of this situation, see below. The line intersects the circle in two points when -4/3 < a < 3/4, otherwise the line and the circle do not intersect. 1
1 1 1 5 -10 -5 5 10 15 -5 3. (i) Construct the mid-point G of AB by using Construction 1.4 from the lectures. Erect a perpendicular to AB at A using Construction 1.5. Draw a circle centre A through G. Let C be the intersection of the circle with the perpendicular. Construct D in the same way: erect a perpendicular to AB at B; draw a circle through G centre B; intersect the perpendicular and the circle to construct the point D. Draw a line CG. Draw a line AD. Construct E as the point of intersection of CG and AD. Use Construction 1.5 to drop a perpendicular to AB from E. Let F be the intersection of this perpendicular and AB. (ii) The triangles AEF and ADB are similar, so there is a constant a such that a|EF| = |DB| and a|AF| = |AB|. The triangles EFG and CAG are similar, so there is a constant b such that b|EF| = |AC| and b|FG| = |AG| = 1|AB|. Since ABDC is a rectangle, |DB| = | AC| and so a|EF| = |DB| = | AC| = b|EF|, and so a = b. Now | FG| = 1|AF|, since |AF| = 1|AB| = 12b| FG| = 2|FG|. So 4|AB| = 2|AG| = (|AF| + |FG|) = 22|AF| = |AF|, as required. 2
4. (i) Draw a circle C1 through B,