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### Apollonius' Problem

```
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
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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