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 withinside circle B. Letís 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 (weíll 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.) Letís 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, itís 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, letís 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 letís 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 |BEí|-|AEí| 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 Eí while animating point C.