EMAT 6680 :: Assignment #7 :: Clay Kitchings

Circle Tangents

In this
exploration we shall explore how to construct a circle tangent to two other
given circles.

First, we will
construct line BD where segment BD is a radius of c2. Next, we will construct point E on line DB such that DE =
AC. Following that we shall
construct the perpendicular bisector of segment AE and label its intersection
with line DB as point F.

Now construct
circle 3 (c3 – in black) centered at point F with radius FD. This circle is tangent to both c1 and
c2.

Note that there is
now an isosceles triangle AFE, with leg segments FE and AF.

Discussion: We
marked off point E outside of c2. What would happen if we chose segment DE so
that point E is inside c2? I will
add this to the figure and we shall proceed as we did earlier.

Notice that using
the same procedure with point E (or point EĠ as it is labeled) on the inside of
c2 produces another circle (c4 – in pink) that is tangent to both the two
given circles as well as the constructed tangent circle.

Consider the locus
of the midpoint of the segment that formed the base of the key isosceles
triangle in the construction. This problem now contains three possible cases
for the two given circles:

Case 1: One circle
is completely inside another circle.

Case 2: The two
circles intersect each other.

Case 3: The two circles
have no points in common (either on the circles or inside the circles).

Case 1: One circle
is completely inside another circle.

Point F is traced
as point D moves around c2. At
first glance we conjecture that this locus (in blue) is perhaps an
ellipse. If it is an ellipse, then
the sum of m(BF) + m(AF) would have to be constant. As visual support for this conjecture (though not a
mathematical proof), we can use the measure function in GSP to check to see if
the sums are measured as constant within GSP as point D is animated around
c2. The results for this
particular figure support the conjecture. For a GSP file, click here.

To prove this,
consider BD which is always constant. We can write BE – ED = BD. Also, BF
+ FE = BD BF + AF =
BD (substituting AF for FE since triangle AFE is isosceles). Therefore the locus is an ellipse since
BD is constant.

Case 2: The two
circles intersect each other.

An analogous
argument can be made for this locus to also be an ellipse as it was in Case 1.

Case 3: The two
circles have no points in common (either on the circles or inside the circles).

It now appears
that the locus of points forms a hyperbola. In order to show this, we must show the DIFFERENCE of BF and
FA is constant, i.e., BF – FA = constant.

We still know that
EF = FA since triangle AEF is isosceles. Let us subtract EB from both sides,
shall we???

EF – EB = FA
– EB BF = FA
– EB BF –
FA = - EB, so therefore BF – FA is constant, implying that we have a
hyperbola.