**EMAT
6690**

**Ken
Montgomery**

**The
Three Tangent Circles Problem**

**Problem Statement:**
Given two
line segments, construct externally tangent circles with these segments
and
construct a common external tangent line. Then construct a circle
tangent to
the two given circles having the same external tangent line. Find the
radius, *c*,
of the new circle in terms of the radii *a*, and *b*, of
the first
two circles.

We
begin with
arbitrary line segments *a*, and *b* and construct the
green circle
with center *B* and radius, *b*. We construct _{} and then we construct
the red, dashed circle, with center *C*, but with radius, *a*.
Notice
that the intersection of Circle *C* and_{}
is *A*, where we construct the red circle *A*,
which is tangent to circle *B* (Figure 1).

**Figure
1:** Arbitrary line segments, *a* and *b*, with tangent
circles *A*
and *B*

If
we consider the
segment, _{}to
be the hypotenuse of a right triangle, then the length of
the base would correspond to the horizontal distance between *A*
and *B*.
The problem now becomes the construction of the vertex *F*,
corresponding
to the right angle of _{}.
We know that the height of the triangle should be equal to *b*
– *a*, and that the length of the hypotenuse is equal to *b + a*.
Assume _{}and_{},
then _{} represents the base
of the triangle and we have Equation 1 from the Pythagorean theorem.

**Equation
1: **_{}

Solving Equation 1 for_{}yields
Equation 2.

**Equation
2: **_{}

Multiplying out the squared terms yields Equation 3.

**Equation
3: **_{}

Distributing the minus sign gives Equation 4.

**Equation
4: **_{}

Combining like terms provides Equation 5.

**Equation
5: **_{}

Taking the square root of both sides gives Equation 6.

**Equation
6: **_{}

Simplification yields
Equation 7,
which gives the horizontal distance in terms of *a* and *b*,
between
Circles *A* and *B*.

**Equation
7: **_{}

To construct the result
of Equation
7, we first find the midpoint of_{}and
construct a circle, with this segment as its diameter. A
line, perpendicular to_{}and
its intersections with the circle are then constructed.
By similar triangles, the orange segment,_{}is
a construction of the desired length, _{}(Figure
2).

**Figure
2:** Construction of _{},
given by the orange segment, _{}

Since
the base of_{}corresponds
to segment_{}and
the height of the triangle should have the value *b*
– *a*, we construct a circle, with center *A* and radius
congruent to_{}.
We also construct a circle, centered at *B*, but with
radius *b* – *a*. The intersection, F, of these two circles
is the
third vertex of our right triangle (Figure 3).

**Figure
3:** Construction of _{}

At
this point, we wish to construct a line, tangent to Circles *A*
and *B*.
Since we know that the line will be tangent to each circle and also a
distance,
*a* (the radius of Circle *A*), from the respective
vertices, *A*
and *B*, we know that the line will be parallel to_{}.
To simplify figure four, we hide part of our previous
construction and proceed to construct a line, parallel to_{},
through point *G* (Figure 4).

**Figure
4:** _{}is
tangent to _{}and_{}

** **

**Part 2: Derivations
of radius, c
and distance,_{}
as a function of radii a, and b**

We
will next wish
to construct a third circle, externally tangent to_{}and_{}and
tangent to_{}.
If we now consider the segment, _{}to
be the hypotenuse of a right triangle, then the length of
the base would correspond to the horizontal distance between *B*
and *C*.
We know that the height of the triangle should be equal to *b* – *c*,
and that the length of the hypotenuse is equal to *b + c*. Assume
_{}and_{},
then _{}represents
the base of the triangle and from the Pythagorean
theorem we have Equation 8.

**Equation
8:**_{}

Solving Equation 8 for_{},
we have Equation 9.

**Equation
9:**_{}

Squaring the binomial terms yields Equation 10.

**Equation
10:**_{}

Simplifying Equation 10, provides Equation 11.

**Equation
11:**_{}

Taking the square root of both sides, gives Equation 12.

**Equation
12:**_{}

Simplification allows us to obtain Equation 13.

**Equation
13: _{}**

We
have an
expression representing the horizontal (*x*-axis) distance between
*B*
and *C*, but it is in terms of the radius, *c*, which we
still do not
know and now seek to determine. Again, using the Pythagorean theorem,
we obtain
Equation 14.

**Equation
14:**_{}

Solving Equation 14 for_{},
we obtain Equation 15.

**Equation
15:**_{}

Squaring the binomial terms yields Equation 16.

**Equation
16:**_{}

Distributing the minus sign and simplifying gives Equation 17.

**Equation
17:**_{}

Taking the square root of both sides, gives Equation 18.

**Equation
18:**_{}

Further simplifying and
solving
Equation 18 for_{}
yields Equation 19.

**Equation
19:**_{}

We restate a previous results: Equation 7 here as Equation 20 and Equation 13 as Equation 21.

**Equation
20:**_{}

**Equation
21: _{}**

Substitution of Equations 20 and 21 into Equation 19 results in Equation 22.

**Equation
22:**_{}

Division of both sides by 2 yields Equation 23.

**Equation
23:**_{}

Rewriting the left side of the equation, we have Equation 24.

**Equation
24:**_{}

Factoring the left side, we obtain Equation 25.

**Equation
25:**_{}

Solving Equation 25 for_{},
provides Equation 26.

**Equation
26:**_{}

Taking the square root of both sides, we have Equation 27.

**Equation
27: _{}**

Substitution of this
result into
Equation 13 yields the expression for_{},
in Equation 28, below.

**Equation
28: _{}**

Simplifying this equation yields Equation 29.

**Equation
29: _{}**

Now that we have
obtained
expressions, in terms of *a* and *b*, for _{} (the horizontal
distance between *B* and *C*), and *c*, the radius of
_{},
we return to our problem of constructing a third circle, _{}tangent
to_{}and
externally tangent to_{}and
_{}.

In our construction, we
will make
use of the relations provided in Equation 13 and Equation 27. Since
Equation 13
depends upon the value of *c*, we begin with a construction of
this value.
Equation 27 is restated below as Equation 30.

**Equation
30: _{}**

Squaring the binomial in the denominator and applying the commutative property of addition, we obtain Equation 31.

**Equation
31: _{}**

Dividing both sides of
the equation
by *a*, yields the ratio given in Equation 32.

**Equation
32: _{}**

Each of the terms in
this ratio is
easily constructible, via similar triangles. First, we construct_{},
with radius *a + b*. Then we construct_{},
with length, _{} (Figure 5).

**Figure
5:** Construction of_{}

We next hide_{}and
construct_{},
with radius_{}.
We wish to construct an isosceles triangle, with base, *b*.
So we next construct_{},
with radius, *b*. We then construct the intersection
of_{}and_{}at
*J *(Figure 6).

**Figure
6:** Construction of_{}

We next hide_{}and_{}and
construct another circle with center, *G* and
radius, *a*. We construct and connect the intersections of this
circle and
sides_{}and_{},
to obtain_{},
which, by Equation 32 and similar triangles, has length, *c*
(Figure 7).

**Figure
7:** Construction of_{},
with length equal to *c*

We next make use of Equation 13, restated below in Equation 33.

**Equation
33: _{}**

This is a relatively
simple
construction that we have already done for ** _{}**,
previously. We construct a right triangle, inscribed in
a circle, with hypotenuse equal to

**Figure
8:** Construction of_{}

In Figure 8, we have
constructed
the circle with center, *K* and radius, *b*. We then
constructed _{}and_{},
which has length, *b + c*. We then construct the
midpoint, *P*, of _{}and_{}.
A line, perpendicular to_{}is
constructed at *K*, the point separating the segments
equal to *b* and *c*, respectively. Then, we construct_{},
which has length equal to ** _{}**.
Next, we hide some of this construction and construct a
circle, with center,

**Figure
9:** Construction of_{},
with radius equal to_{}

In Figure 9, we also
constructed a
line, perpendicular to_{}at
*Q*. We next (Figure 10), construct a circle, with
center, *Q* and radius equal to *c* (congruent to_{})
and its intersection (*R*) with the perpendicular
line.

**Figure
10:** Construction of_{},
with radius, *c*

At this stage, we hide
unnecessary
parts of the construction and construct the third, externally tangent
circle (_{}),
with radius, *c*, which is a horizontal distance,_{}from
the point, *B* (Figure 11).

**Figure
10:** Construction of three, externally tangent circles and a
mutually
tangent line

Open
the file, ThreeTangentCircles.gsp
to explore this
construction and verify that it is valid for different values of *a,*
and *b*.

**Alternate statement
of Equation
27**

Although our derivation of formulas culminated in Equations 27 and 29 (below), Equation 27 is rewritten, in a more aesthetic form.

**Equation
27: _{}**

**Equation
29: _{}**

We begin with Equation 27, given here as Equation 34.

**Equation
34: _{}**

Solving Equation 34 for
** _{}**,
we have Equation 35.

**Equation
35: _{}**

Squaring the binomial term yields Equation 36.

**Equation
36: _{}**

Applying the distributive property, we obtain Equation 37.

**Equation
37: _{}**

Factoring the right side, we have Equation 38.

**Equation
38: _{}**

Taking the square root of both sides yields Equation 39.

**Equation
39: _{}**

We next divide both
sides of the
equation by** _{}**(Equation
40).

**Equation
40: _{}**

This division results in Equation 41.

**Equation
41: _{}**

Applying the commutative property of addition to the right side, we obtain Equation 42.

**Equation
42: _{}**

Derived results are summarized below.

_{}

**and**

_{} or _{
}

_{}