Circumscribing Tangent CirclesDate: 04/13/99 at 09:20:13 From: s day Subject: Problem of Apollonius My geometry class read about this problem in a vignette about Apollonius: Given three circles of various sizes in a plane, circumscribe a circle about them. We tried straightedge and compass and the Geometer's Sketchpad, but we haven't had any success. Date: 04/29/99 at 03:27:49 From: Doctor Floor Subject: Re: Problem of Apollonius Hi, S. Day, Thanks for the fascinating problem. The construction problem turns out to be not easy at all, so it is no surprise that you have not succeeded in doing this construction in class. There is, however, a simpler case of this problem: when the three given circles of various sizes in the plane are mutually tangent. I will show you the solution of that more simple problem. To solve it we will make use of "inversion" in a circle: Let P be a point, and C a circle with center O and radius r. Then the inverse of P in the circle is the point P' on the line OP, such that OP*OP' = r^2. Geometrically this inverse P' is found as follows when P is outside the circle: Construct a tangent from P to the circle. Let X be the point where this tangent meets the circle; then P' is the foot of the perpendicular altitude from P' to OP. When P is inside the circle, you just work the other way around. When P is on the circle then P' = P. Here's an example: The following properties of inversion will be used: - Each point on the circle used as base for the inversion is fixed, - Straight lines through O are inverted to themselves, - Circles passing through O are inverted to straight lines not through O ("circles with infinite radii") and vice versa, - Other circles are inverted to circles, - When circles are tangent, their inverted circles are tangent too (also when they have become circles with infinite radii). We will now do the construction as shown in the following picture: We start with three mutually tangent circles, with centers A, B, and C respectively. Let F be the point where the circles centered at B and C meet. We construct line FA. The first point from F where FA meets the circle with center A is D. Our base circle will be the circle through D with center F. This circle is drawn in red, and is tangent to the A-circle. The inverses of the B- and C-circles are easily found, because we know they will be straight lines. Their intersections with the red circle are fixed. So if we connect those, we find the two parallel straight lines that are the inverses of the B- and C-circles. The A-circle is inverted to a circle tangent to the A-circle and the red circle in point D, and also tangent to the two lines we just constructed. Its center can be found easiest by intersecting FA with these two lines, and then constructing the midpoint of the two intersections. Now we are going to construct the inverse of the circle we are looking for. It must be tangent to the two parallel straight lines and the circle, the three inverted circles we just constructed. This is a translation ("shift") of the inverted A-circle, and is easily found. In the picture this is the green circle. We find that this green circle meets the three inverted circles in three points (W,Q and A' in the picture), which must be the inversions of the three tangent points with the circle we are looking for. So we should invert these points back. We can easily to that. As an example: the inverse of W must be on the line WF, and also on the C circle, and of course the outside point in this circle. So we find the three tangent points (in picture: Y, S and C') of the circle we have to construct and the three original circles. Finally, we find the desired circle as the circumcircle of YSC. As you can see, inversion is a powerful tool here. However, I am afraid that it is not sufficiently powerful to do the more general problem of Apollonius you described. At least, I don't see how. I hope your class will enjoy this construction anyhow. If you have trouble finding the three circles for the starting position: start with a triangle ABC. Construct its incircle, and let D on BC, E on AC and F on AB be the points where the incircle meets the sides. The circles with center A through E and F, center B through D and F and center C through D and E are three mutually tangent circles. If you have a math question again, please send it to Dr. Math. Best regards, - Doctor Floor, The Math Forum |
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