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### Apollonian Construction Problem

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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

I hope this helped, but if you need more, just write back.

Best regards,
- Doctor Floor, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
College Constructions
High School Constructions

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