Date: 09/07/2000 at 15:01:26 From: Jeroen Cottaar Subject: Making a circle that intersects three others Given three non-intersecting circles, is it possible to construct a circle that intersects each of them in one point using only a compass and straightedge?
Date: 09/08/2000 at 11:09:56 From: Doctor Floor Subject: Re: Making a circle that intersects three others Hi, Jeroen, Thanks for your question. If I understand you correctly, you are asking for the solution of Apollonius' Problem, which asks for a circle tangent to three given circles. Generally there are eight solutions to this problem. In the Dr. Math archives you can find one solution to only a particular case of Apollonius' problem: * The constructed circle circumscribes the three given ones (the three circles are all tangent to the constructed one internally), * The three given circles are each tangent to the other two. The solution is here: Circumscribing Tangent Circles http://mathforum.org/dr.math/problems/day.04.13.99.html The constructed circle in this special case is usually referred to as the outer Soddy circle. The reason I point you to this case first is that the message also explains inversion. At the moment I wrote the above reply I didn't realize I could use inversion for your more general problem. With inversion, however, we can find two solution circles to the more general problem in the following way: Let (C1), (C2), and (C3) be three non-intersecting circles, with centers C1, C2, and C3. Let's assume that the radius of C1 is the smallest one. Then we can reduce the problem by subtracting from the radii of the three circles the radius of C1 (and consequently also subtracting this amount from the radius of the circle to be constructed). The result is a point C1, and two smaller circles, [C2] and [C3]. Let's call the circle that we have to construct (C). Now if we perform an inversion in any circle with center C1 - for instance (C1) - then, because (C) passes through C1, the image of (C) will be a line l. Also, the images (C2') and (C3') of [C2] and [C3] must be tangent to this line l. That gives the following construction: 1. Construct the reduced circles [C2] and [C3]. 2. Construct the inverses (C2') and (C3') of [C2] and [C3] with (C1) as base for the inversion (or any other reasonable circle with C1 as its center. Hint: it is easier to construct inversions of circles intersecting the base circle. 3. Construct the two tangent lines to (C2') and (C3') that have (C2') and (C3') on the same side of the line. For a construction aid on this, see from the Dr. Math archives: Line Tangent to Two Circles http://mathforum.org/dr.math/problems/cruz07.01.99.html 4. Invert these lines, in (C1), to two circles (through C1). 5. One of the image circles - the greater - will be circumscribing [C2] and [C3]. To the radius of that image circle you can add the radius of (C1), yielding one solution of the problem, which has (C1), (C2) and (C3) inside the constructed circle. The other image circle can be used as well. It kisses [C2] and [C3]. If now you subtract from the radius of this image circle the radius of (C1) then this solution circle will be "enclosed" by the three given circles, so (C1), (C2) and (C3) will all be outside the constructed circle. Those are only two solutions to a problem I said had eight solutions. The other solutions are found by reducing the given circles in a different way. Instead of subtracting equal amounts from all three radii of the given circles, one can also subtract from two and add the same amount to the radius of one. In doing so one can construct the different possibilities with one or two of the given circles inside the constructed circle and the other one(s) outside. I hope this will help. If you need more help, just write back. Best regards, - Doctor Floor, The Math Forum http://mathforum.org/dr.math/
Date: 02/21/2002 at 09:52:05 From: Bruno Folens Subject: Tangent circles How to find the other four solutions of circle-circle-circle ? Bruno Folens
Date: 02/22/2002 at 04:09:22 From: Doctor Floor Subject: Re: Tangent circles Hi, Bruno, In this situation, the circle-circle-circle situation is transformed into circle-circle-point by subtracting equal amounts from all radii. This gives two apollonian circles. Instead of subtracting from all, one can add the amount to one and subtract from the others. That gives three more translations into circle-circle-point, each yielding two more solutions. From there we come to eight solutions. Here is an example of the three new circle-circle-point situations: If you have more questions, just write back. Best regards, - Doctor Floor, The Math Forum http://mathforum.org/dr.math/
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