Apollonian Construction Problem
Date: 03/06/2001 at 13:40:12 From: Zelma Knight Subject: Geometry - Apollonian Construction Problem Given a line and two points A and B, tell me how to construct a circle containing the two points and tangent to the line. I know that if points A and B lie on the circle they are equi-distant from the point called the center of the circle (we'll call this point C), and that point C lies on the perpendicular bisector of chord AB. How do I find the point of tangency to the line?
Date: 03/06/2001 at 15:59:22 From: Doctor Floor Subject: Re: Geometry - Apollonian Construction Problem Hi, Zelma - thanks for writing. There are several ways to do this construction. Here is one that we can derive ourselves. In a rectangular grid we suppose that the given line is the x-axis, and we take A to be the point (0,1). Furthermore, we let y = ax+b be the perpendicular bisector of AB. The center of the circle must be on this line, and this center must be as far from A as from the x-axis. Let (p,ap+b) be the coordinates of the center; this means that: distance from x-axis = ap+b distance from A(0,1) = sqrt[p^2 + (ap+b-1)^2] Those two being equal gives: (ap+b)^2 = p^2 + (ap+b-1)^2 a^2p^2 + 2abp + b^2 = p^2 + a^2p^2 + 2a(b-1)p + b^2 - 2b + 1 0 = p^2 - 2ap - 2b + 1 With the quadratic equation this gives for the value of p: p = a +/- sqrt(a^2 + 2b - 1) How can we construct this? Let me give you some guidelines: Part I: 1. Length a can be found in the following way: find the y-value of the perpendicular bisector of AB for x = 1. This is y = a+b. The vertical difference with the y-intercept is the requested length. Note that this might be negative when the perpendicular line is descending. 2. The y-value of the perpendicular bisector for x = a is then y = a^2+b. Here is a figure of this part of the construction; in it the final point (a,a^2+b) is called Q: Part II: Add to the y-value of this point b-1, which is exactly the distance from A to M. Mark the y-value, which is a^2+2b-1, on the y-axis, using point R. Draw the circle with diameter OA or OR, whichever is larger (in the case of our figure, this is OR). From point A or R in the interior, draw a line perpendicular to OR, and let it intersect the circle at point U. Then OU has diameter sqrt(a^2+2b-1). The truth of this can be seen from the fact that in a right triangle OUR, with altitude OA, the following identity holds: OA*OR = OU^2 (which can be proven by use of similarities). Here is a figure: Part III: I leave this for you: you should draw the circle with center S(a,0) (which is on the x-axis below Q) and diameter OU to find the x-values of the circle centers. From there, go to the perpendicular bisector of AB and find the two circle centers - yielding the two solutions for your construction problem. I hope this helped, but if you need more, just write back. Best regards, - Doctor Floor, The Math Forum http://mathforum.org/dr.math/
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