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### Circumscribing Tangent Circles

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Date: 04/13/99 at 09:20:13
From: s day
Subject: Problem of Apollonius

Apollonius:

Given three circles of various sizes in a plane, circumscribe a

We tried straightedge and compass and the Geometer's Sketchpad, but we
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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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Associated Topics:
High School Conic Sections/Circles
High School Geometry

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