TANGENT CIRCLES
By Dario Gonzalez Martinez
The problem is as follow:
Given
two circles and a point of one of the circles, construct a circle tangent to
the two circles with one point of tangency being the designated point”
We will begin considering that the two given circles are such that one is completely inside of the other. So, we should consider that there exist two cases for this problem under the condition before mentioned. These cases problems are presented in figure 1 below:
Tangent circle does not surround the inside circle 
Tangent circle surrounds the inside circle 
Figure 1(a) 
Figure 1(b) 
Green circles are the given circles, and points E and D are the given points of tangency for each figure, respectively. Then, the blue circle is the tangent circle sought.
We will analyze each case separately.
TANGENT CIRCLE DOES NOT SURROUND
THE INSIDE CIRCLE
I. Tangent circle construction
Consider figure 2 below:
Figure 2 
Here (figure 2) are the two given circles (circles with centers at A and B, respectively) and the given tangency point C. The center of the tangent circle must lie on line AC.
First, we should prove why the tangent circle’s center lies on line AC. Since segment AC is a radius of circle with center at A, a tangent line j to this circle that passes through C must be perpendicular to segment AC. Let a point T be the center of the tangent circle at C, and let segment TC be the radius of this circle as follow in figure 3 below:
Figure 3 
Since TC is the radius of the tangent circle at C, line j must also be perpendicular to segment TC. So, line AC and segment are either parallel to each other or the same line. However, line AC and segment TC intersect each other at C, so they are not parallel to each other. Thus, line AC and segment TC must be the same line, so point T lies on line AC.
Now we can go ahead with the tangent circle construction. Suppose that we already know the location of the tangent circle’s center T, and consider a circle with center at C and the same radius that the circle with center B as follow below:
Figure 4 
Since the circle with center at T is tangent to both circles with centers A and B, respectively, the distances TE and TC are equal to each other. Also, the distances TB and TF are equal to each other by construction, so triangle BTF is isosceles with base BF. Therefore, the perpendicular bisector of BF must passes through T.
Now we have the construction to find the tangent circle center, which I summarize as follow:
1) Construct a circle with center at C and radius equal to the radius of the circle with center B.
2) F is intercept between circle constructed in 1) and line AC.
3) Construct the segment BF.
4) Construct the perpendicular bisector of BF.
5) The intersection between the perpendicular bisector and line AC is the tangent circle’s center T.
6) Construct the tangent circle with center at T and radius TC.
Figure 5 shows the construction done.
Figure 5 
II. Discussion about the locus
of the center T
Since point C is an arbitrary point on circle with center A, there are several possible locations for tangent circle’s center T. So, let’s move the point C around the circumference of circle with center A, and we will observe the trace of the center T. The sequence shown below suggests the shape of the center T locus:
Sequence 1 
This seems the focus of tangent circle’s center is an ellipse with foci at the centers of the given circles. If you want to see an animation in Geometer’s Sketchpad (GSP) click here. It just remains to show that this locus is effectively an ellipse. To this end, consider the figure 6 below:
Figure 6 
We can observe that segments TE and TC are equal to each other. So, the sum TB + TA is equal to the sum between the radii of the given circles. Thus, the sum TB + TA is constant, and the locus of tangent circle’s center T is an ellipse by definition.
Now we raise the reasonable questions:
· What happens if the circle with center B is tangent to circle with center A, but it is still inside?
· What happens if the circle with center B intersect circle with center A?
· What happens if the circle with center B is tangent to circle with center A, but it is outside?
· What happens if the circle with center B does not intersect circle with center A and is outside?
We will answer each question separately.
The circle with center B is tangent to circle with center A, but it
is still inside
We should take a look at the sequence 2 below:
Sequence 2 
Now the given circles are tangent each other at E, and the sequence suggests an ellipse for the locus of tangent circle’s center T once again (click here for an animation in GSP). It just left to show that the mentioned locus is indeed an ellipse with foci at the centers of the given circles. To that end, consider the figure 7 below:
Figure 7 
From the figure 7 above we can observe that TF = TC, so the sum TB + TA is constant since it is the sum between the given circles’ radius. Thus, the locus is an ellipse by definition again.
The circle with center B intersect circle with center A
Observe the following sequence:
Sequence 3 
One more time the sequence suggests that the locus for the tangent circle’s center T is an ellipse with foci at the centers of the given circles (click here for an animation in GSP). We should just show that focus is indeed an ellipse, so consider the figure 8 below:
Figure 8 
From the figure 8 above we can observe that TE = TC, so the sum TB + TA is constant since it is the sum between the given circles’ radius. Thus, the locus is an ellipse by definition again.
The circle with center B is tangent to circle with center A, but it
is outside
For this case it is unnecessary to observe a sequence; instead, we will consider the following figure:
Figure 9 
Since we are in the case where the tangent circle does not surround the circle with center B, and considering that the tangent circle must be tangent at D and C, it cannot be located such that it surrounds circle with center A.
First, we will consider that points D and C are different. Let T be the tangent circle’s center, and let TC and TD radius of this circle. If we consider the tangent lines m and l to the circle with center A, then AD and AC are perpendicular to lines m and l, respectively. Since tangent circle is tangent at D and C, lines m and l are also perpendicular to TD and TC, respectively.
So, segments TD and AD are either parallel to each other or they are the same segment. Since TD and AD share point D, they are not parallel to each other; therefore, they are the same segment. Thus, T lies on AD. By a similar reasoning we can conclude that T also lies on AC, so T lies on the intersection between AD and AC. Thus, T coincide with A which means that A is the center of the tangent circle, which in turn means that the tangent circle coincide with the circle with center A.
Since the tangent circle and the circle with center A are the same, the locus of tangent circle’s center is a point, the point A.
On the other hand, if we consider that D and C are the same point, then there are infinite tangent circles to the given circles. All these tangent circles will have center on the ray DA as follow in figure 10 below:
Figure 11 
This time the locus of tangent circle’s T_{i} center will be the ray DA for all i.
The circle with center B does not intersect circle with center A and
is outside
It is precise to observe a sequence as follow again:
Sequence 4 
This sequence suggests a hyperbola for locus of the tangent circle’s center T with foci at the centers of the given circles (click here for an animation in GSP). It just remains, once again, to prove that the mentioned locus is a hyperbola. To this end consider the figure 11 below:
Figure 11 
We should observe that TE = TC always since they are radius of the tangent circle. Therefore, if we rest the distance TA from TB we will obtain AC + BE always; thus, the difference between TB and TA is constant and equal to the sum of the given circles’ radius. Therefore, the locus is an hyperbola by definition.
III. Tangent circle construction
for other cases
Although we analyze the locus of the tangent circle’s center T for different cases, we did not discuss the construction of the mentioned tangent circle when the given circles have different relative position from the assumed positions for our first construction explanation. That is why we should discuss the construction for two more cases: intersected given circles and disjoint given circles.
Intersected given circles
We will deduce the construction for this case by using the same strategy that we use in section I, that is, we will assume the problem was solved. So, we should observe figure 12 below where the tangent circle already appears there:
Figure 12 
The given circles are the circles with centers at A and B, respectively. T is the tangent circle’s center, and let circle with center C a circle with the same radius as circle with center B. Then TB = TF, and triangle BTF is isosceles with base BF. Thus, the perpendicular bisector of BF must pass though the tangent circle’s center T, and we have our construction which I summarize as follow:
Given two intersect circles with centers at A and B, respectively and a point C on one of the circles (this is the tangency point desired), then
1) Construct the line AC.
2) Construct a circle with center C and radius as the radius of circle with center B.
3) F will be the interception between the circle with center C and line AC as shown in figure 12 above.
4) Construct the segment BF.
5) Construct the perpendicular bisector of BF.
6) The tangent circle’s center will be the intersection between the perpendicular bisector and line AC.
7) Construct a circle with center T and radius TC.
Disjoint given circles
We can deduce the construction for this case by applying a similar reasoning as that used in the case previously presented. Actually, we could use the same instruction as above but considering the figure 13 below:
Figure 13 
The only given information are the yellow circles and the point C. We will construct the blue circle (tangent circle) by following the same instruction above mentioned.
TANGENT CIRCLE SURROUNDS THE
INSIDE CIRCLE
As I mentioned before, there is another solution for the problem of finding the tangent circle to two given circles in a given point C. This solution involve a tangent circle that surround the circle inside of the other given circle (we begin our analysis with this assumption).
I. Tangent circle construction
We will assume, again, the problem is solved, so we can deduce the construction by analyzing the geometric properties involved in it. We should consider figure 14 below:
Figure 14 
The given circles are the circles with centers at A and B, respectively, and the given point of tangency is point C. Suppose point T is the tangent circle’s center, and let circle with center C be a circle with the same radius that circle with center B. Thus, TB = TF and triangle BTF is isosceles, so the perpendicular bisector of BF passes through point T, and we obtained our construction.
Figure 15 shows how the construction should look like. Given two circles with centers A and B and a point of tangency C, as figure 15 shows, the construction is the following:
1) Construct the line AC.
2) Construct a circle with center C and radius as the radius of circle with center B.
3) F will be the interception between the circle with center C and line AC as shown in figure 15 below.
4) Construct the segment BF.
5) Construct the perpendicular bisector of BF.
6) The tangent circle’s center T will be the intersection between the perpendicular bisector and line AC.
7) Construct a circle with center T and radius TC.
Figure 15 
II. Discussion about the locus of the center T
Now we will discuss the locus of the tangent circle’s center T since point C is arbitrary. At the same time, we should consider the different cases for the relative position of the given circles (one inside, tangent, intercepted, disjointed) and discuss the locus of tangent circle’s center in each case. It is possible to elaborate an analysis similar to that done in the first solution for the problem. In this manner we will have the following cases and locus:
One given circle is inside the other one 

The trace of point T suggests an ellipse with foci at the given circles’ centers. This is actually the case since if you consider the circle with center C and radius equal to the radius of the circle with center B, you can note that TA + TB is constant and equal to the difference between the given circles’ radii. Indeed, TA + TB = AC – CD since TB = TD. So, the red trace is an ellipse by definition (click here for an animation in GSP). 

One given circle is tangent to the other one, but it is still inside 

Since the tangent circle surrounds the circle with center B, and considering that the tangent circle must be tangent at D and C which are considered different points, it cannot be located such that it surrounds circle with center A. By considering a similar argument that presented below figure 9, we conclude that the tangent circle and the circle with center A must coincide. Then the locus for the tangent circle’s center T is the point A. On the other hand, if points C and D coincide, there are infinite solutions for the problem, and the locus for the tangent circle’s center is the ray EA. 

The given circles intersect each other 

This time the trace of center T suggests hyperbola with foci at the given circles’ centers as the locus. Consider the circle with center C and radius equal to the radius of the circle with center B. We have that TB = TD, and we also note that TB – TA is constant since TB – TA = TD – TA = AD, which in turn is the difference between the given circles’ radii. Thus, the locus of the tangent circle’s center is a hyperbola by definition (click here for an animation in GSP). 

One given circle is tangent to the other one, but it is outside 

Once again the locus suggested by the trace of T is a hyperbola with foci at the give circles’ centers. If we apply the same reasoning presented above, we would conclude that TB – TA = TD – TA = AD, which in turn is the difference between the given circles’ radii. Thus, the locus of the tangent circle’s center is a hyperbola by definition (click here for an animation in GSP). 

The given circles are disjointed 

Finally, the trace of T suggests a hyperbola with foci at the given circles’ centers as the locus. In the same manner that we already showed it before, we can note here that TB = TD (since circle with center C has the same radius that circle with center B), so TA – TB = TA – TD = AD is constant since it is the difference between the given circles’ radii. Thus, the locus is a hyperbola by definition (click here for an animation in GSP). 
GENRALIZATION
FOR CONSTRUCTIONS
The construction of a tangent circle to two given circles and a given tangency point has almost the same structure or steps for both solutions (tangent circle surrounds or not the circle with center B). The only difference lies what you consider as the interception between the constructed auxiliary circle with center C and the line AC. I will explain it much clearly, so we will consider the steps for the construction mentioned below figure 4 and figure 14:
1) Construct the line AC.
2) Construct a circle with center C and radius as the radius of circle with center B.
3) F will be the interception between the circle with center
C and line AC.
4) Construct the segment BF.
5) Construct the perpendicular bisector of BF.
6) The tangent circle’s center T will be the intersection between the perpendicular bisector and line AC.
7) Construct a circle with center T and radius TC.
We should consider the figure 16 below:
Figure 16 
T2 and T1 represent the center of the tangent circle that surrounds and not the circle with center B, respectively. The auxiliary circle with center C and radius equal to the radius of circle with center B intercepts line AC twice at D and E. Step number 3 asks you to join one of these intersections with point B. This segment will be the base of an isosceles triangle. If you want the blue tangent circle, you must construct the segment BE; on the other hand, if you want red tangent circle, you should construct segment BD.
Suppose that you want the blue tangent circle. I drew the center of this circle, but you really do not know where it is. However, you know that T1B = T1H + (the radius of circle with center B). Points T1, H and B are in the same line for the same reason that I explained in the first construction for the first solution. T1B is one side of the isosceles triangle in which we base our construction; we just need to find the third vertex of it (since B and T1 are two of them, even though you do not know where T1 is). Since T1H = T1C, we need to ADD the radius of the circle with center B to T1C, and we will find the third vertex needed. That is why, if you want the blue tangent circle, you choose E as the intersection between circle with center C and line AC.
On the other hand, suppose that you want the red tangent circle. I also drew the center of this circle, which, of course, you do not know where it is either. However, you know that T2B = T2I  (the radius of circle with center B). Points T2, I and B are in the same line for the same reason that I explained in the first construction for the first solution. T2B is one side of the isosceles triangle in which we base our construction; like before, we just need to find the third vertex of that isosceles triangle. Since T2I = T2C, we need to REST the radius of the circle with center B to T2C, and we will find the third vertex needed. That is why, if you want the red tangent circle, you choose D as the intersection between circle with center C and line AC.
Thus, the steps for the construction of a circle tangent to two given circles at a given point for one of them are the following:
1) Construct the line AC.
2) Construct a circle with center C and radius as the radius of circle with center B.
3) E and D will be the interception between the circle with center C and line AC.
4) If you want a tangent circle that surrounds the circle
with center B, construct the segment BD.
Otherwise, construct segment BE.
5) Construct the perpendicular bisector of BD or BE, according to the case.
6) The tangent circle’s center T will be the intersection between the perpendicular bisector and line AC.
7) Construct a circle with center T and radius TC.
This construction will work since our reasoning do not depend on the position of the two given circles, even though they seem disjointed in figure 16.