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 E’ 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 è 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 à BF = FA – EB à BF – FA = - EB, so therefore BF – FA is constant, implying that we have a hyperbola.