Carrie Dillman

This assignment is an exploration of the problem of being given two circles and a point on one of the circles and to construct a circle tangent to the two given circles with one of the points of tangency being the given point on one of the circles.


My experience with this assignment has been representative of what high school students go through when learning this topic. For starters, I constructed each scenario individually before it was kindly brought to my attention that I only needed one construction. It was just a matter of moving and resizing the circles to obtain the 8 constructions I was after. I don't consider the time spent making separate constructions wasted though. Nor will I consider it time lost or wasted if one of my students takes the same long trip around the mulberry bush. It was through all 8 constructions that I really began to understand what I was looking at, looking for, and the roll of each point, line, and circle in the construction.


The Construction

Let's begin this investigation by doing the construction. First, recall how many ways circles can interact with each other. They can either be disjoint, where they don't touch each other at all. Circles can intersect at two points. Or they can be tangent at exactly one point. We are going to look at the constructions of circles tangent to two given circles in each of the three above scenerios and proceed to explore the behavior of the tangent circles.

So, we are given that there are two circles and a point on one of the circles that will end up being a point of tangency for the tangent circle we are trying to construct. We will call the two given circles, circle A and circle B, and designate the point of tangency as point C. Notice the two points on circles A and B that are not named. Leave them as is and do not hide them. You will use them later to resize your circles.

Now we need to construct line AC. The centers of our tangent circles will lie on this line.

From here we need to sort of work backwards to understand the end result of the construction. Follow me. Here is what we want to end up with.

Wow! We magically have a circle tangent to both circles A and B with the point of tangency at point C and the center of our tangent circle, point E, lies on line AC just as we wanted. Well, certainly not magically. Hmmm. How did we know where to put point E? What do we notice about the radius of the tangent circle? Since circle B is inside the tangent circle, then the diameter of circle B is definitely included in the length of the radius. That might be useful. What if we made a copy of circle B using point C as the center? Construct the circle using C as the center point and the radius of circle B.

Notice I constructed the points where circle C and line AC intersect. Since the radius of the tangent circle is the diameter of circle B plus an unknown length the rest of the way to point E, constructing circle C only gives us half the diameter of B or the radius. No worries though. What we are saying is that the distance from E to C is the radius of the tangent circle just as the sum of the diameter of B and the unknown distance between circle B and E is also the radius of the tangent circle. So in essence we have an isosceles triangle.

What becomes apparent then is that point E, the center of the tangent circle, is the point where the perpendicular bisector of the base of the isosceles triangle intersects line AC.

Consider the isosceles triangle if you "subtract off" the length of the radius of circle B from both legs of the isosceles triangle and construct a new base.

Now the base is the segment from the center of circle B and intersection F of the circle C and line AC. Notice the perpendicular bisector of the new base still intersects line AC at point E. To assure ourselves that E is in fact the center of the circle tangent to the given circles A and B at the point C of tangency, go ahead and construct the circle by center point E and the point on the circle C.

As you can see we are right back to where we started with the tangent circle, only now we know how to find the center and hence the circle. The other point on circle C where line AC intersects it may also be used to find a second tangent circle. The difference in the construction is that the base of the isosceles triangle will be the segment from that point of intersection to point B the center of the small given circle. As in the previous construction, the point of intersection of the perpendicular bisector of the base of the isosceles triangle and line AC is the center of the second tangent circle. This second tangent circle, as you can see, is on the outside of the small given circle, whereas the other tangent circle emcompasses it.

In summary, the center of a circle tangent to two given circles at one designated point lies on the line through the center of the given circle with the designated point and the designated point. The distance from the center of a tangent circle to the center of the other given circle without the designated point is always the sum of the radius of the tangent circle and the radius of the other given circle.


Use my script tool to investigate the circles tangent to two given circles, one smaller than the other, when the two circles are disjoint, intersecting, and tangent to each other.

Now let's take an indepth look the center of the tangent circle. The tangent circles we constructed above are specific for that particular point of tangency. As the location of the point of tangency on the given circle changes, so will the tangent circles and their centers. To illustrate this point I have animated the tangent point so that it will move around the given circle and show the many possible tangent circles. Click here to see.

You didn't happen to notice the path of the center of the tangent circle did you? Let's go ahead and trace that path while the designated point moves along the given circle and see what we find.

Tracing the path of G, the center of the tangent circle, we see that the locus of points is an ellipse with foci, A and B, the centers of the given circles. How do we know this is an ellipse? By the definition of an ellipse, the distance from focus A to G, AG, the center of the tangent circle, plus the distance from focus B to G, BG, is always constant. To prove this is what we have constructed, we need only find another length equal to the sum of those distances that we know for sure is constant. This is easily proved by the construction.

BG + AG = BG + (AD - GD), because AG = AD - GD

BG = GD because they are the sides of the isosceles triangle. So, BG + AG = AD. Well, AD =AC + CD, which just so happen to be the radii of the two given circles A and B. And we know the length of these radii are always constant, therefore BG and AG are always constant and the locus of the tangent circle G is in fact an ellipse.