An
Exploration of Tangent Circles

By: Sydney
Roberts

This
investigation will deal with constructing tangent circles. Given two circles
and a point on one of the circles, these instructions will show how to
construct a circle tangent to the two given circles with one point of tangency.

The first
case we will investigate is constructing a tangent circle when the two given
circles involve one circle completely inside the other as shown below:

Now we want
to construct a line through points A and C. If we want a circle to be tangent
to circle A and circle B that is tangent at the point C, then it makes sense
that the center of our desired circle would lie on the same line as After constructing that line, we also want a
circle centered at C with the same radius as circle B.

Now we want
to label the point where circle C intersect with inside circle B. Let’s call this point D and
then construct a segment from B to D.

At this
point, we want to construct the perpendicular bisector to the segment BD.
Because this perpendicular bisector will not be parallel to the line AC, it
will cross this line. We can label this point of intersection as point E.
Because of the properties of perpendicular bisectors, we then have an isosceles
triangle BED.

Now remember
the goal here is to find a circle that goes through point C and is tangent to
circle A and circle B. So we need a center point and a radius for this desired
circle. If we let |EC| be the radius of our desired circle and the center be
the point E, then the circle would be tangent to Circle A at point C as
desired. Would it also be tangent to circle B? Yes, because |ED|=|EB| because
of the properties of the perpendicular bisector. Also, |DC|=the radius of
circle B. So if we were to extend the segment BE to where it intersects circle
B again (we’ll label this point F). Then |EF|=|FB|+|BE|=|ED|+|EC|=|EC|. Hence,
the circle with radius |EC| centered at E would also be tangent to circle B at the
point F as shown below.

Hiding
everything but our two given circles and the desired circle, the resulting
picture is a little cleaner.

Is this the
only tangent circle that can be constructed? No. Remember when we chose to
label D as the point where Circle C intersected AC inside circle A? (Refer to
the third figure.) Let’s go back and choose the point of intersection to lie
outside of Circle A.

Now we want
to construct segment BD, and again find the perpendicular bisector, labeling
the point of intersection between the perpendicular bisector and line AC as
point E.

Just as
before, notice that we can construct an isosceles triangle BED. We can label
the point where BE intersects circle B as the point F.

Just as
before, we can create a circle centered at E with a radius of |EC| and this
circle will also be tangent to circles A and B at points C and F respectively.

Cleaning up
the picture, we can see the two tangent circles clearly.

The other
two cases are when the given circles overlap, and when the given circles are
disjoint. However, the previous construction can result in a script tool that
will also construct the next two cases. For a link to this GSP script tool,
click here.(Refer
below to see the other two cases)

Now, it’s
interesting to consider the loci of the centers of our tangent circles for the
three different cases. Shown below is the result of what happens when you trace
the loci of the centers of both the tangent circles when the point of tangency
revolves around the first given circle.

The traces
(The thick red) appear to make ellipses with the centers of our two given
circles as the foci. To show this is true, let’s look (again) at the figure
below.

To show that
the inner ellipse (tracing the center of the red tangent circle) is in fact an
ellipse, we need to show that the sum of |BE| and |AE| is a constant.

Well we know
that |AE|=|AC|-|EC| and |BE|=|EF|-|BF|=|EC|-|BF|

Then
|BE|+|AE|=|EC|-|BF|+|AC|-|EC|=|AC|-|BF| But note that AC is the radius of
circle A and BF is the radius of circle F. Hence, these radii remain constant,
and then |AC|-|BF| is constant. Hence, |BE|+|AE| is
constant desired.

To show that
the outer ellipse (tracing the center of the blue tangent circle) is in fact an
ellipse, the same calculations would be done, except this time refer to the
following figure.

Now let’s
consider what the traces would look like if our two given circles were
overlapping.

Here it
appears as if the smaller tangent circle (the blue circle) would make an
ellipse while the larger tangent circle (the red circle) would make a
hyperbola. We have already shown the mathematics involved for proving the
tangent circle is an ellipse, so I will leave this to the reader to show for
this case. However, we have not shown how to prove the path of the loci is a
hyperbola. Well, in order to show this we need to prove that the difference of
AE and BE is a constant. Therefore, consider the image below.

Well, we
know that |BF|=|CD| since these are both radius of the same size circle. We
also know that |EF|=|CE| since this is the radius of the red circle. Then |AE|
= |CE|-|AC| and |BE|=|EF|-|BF|=|CE|-|CD|.

Thus,
|BE|-|AE|=|CE|-|CD|- ( |CE|-|AC|) = |AC|-|CD| and
these distances remain constant because they are the same as the radii of our
given circles. Hence, the difference is constant and the trace of the loci
forms a hyperbola.

For the last
case, consider the circumstance where are two circles are completely disjoint.

Again, these
traces appear to be hyperbolas and it will be left to the reader to show that
the mathematics hold just as they did in the second case. Remember, you need to
show that |BE|-|AE| is a constant and that |BE’|-|AE’| is a constant.

For a GSP
file to play around with these loci, use the script tool that is linked above
and trace points E and E’ while animating point C.