For this assignment we will be exploring the behaviors of tangent circles for three cases:

i. When one given circle lies completely inside the other.

ii. When the two given circles are disjoint.

iii. When the two given circles overlap.

Let's begin our exploration by seeing what happens when one given circle lies inside the other given circle.

Here we have our given circle with Center A inside of our other given circle with Center C. Our construction has given us a circle with Center G tangent to both given circles as desired. But how did we find this tangent circle with Center G and a radius of GB? Let's walk through it together. We are given that one circle lies inside of the other circle so that covers our positioning of circles with Center A and Center C. But to find our tangent circle we must search for an isosceles triangle that will provide the radius of our tangent circle. So our next step is to construct a congruent circle to the circle with Center C. This circle will be positioned such that B is the center and the radius equals CD. We find B by simply selecting a point on the Circle A. We mark B as our center and construct a circle with the radius of Circle C. So we now have another large circle colored blue with center B and radius CD encircling circle with Center A. Next we will construct a line through the points A and B. The intersection of that line and our new circle with Center B will be the point E. Here is where the construction of our isosceles triangles begins. We will construct the segment CE and midpoint of that segment as F. By drawing a perpendicular line to CE through F we will find an intersection point between the line AB and our perpendicular line to CE. That point will be G and serve as a vertex of our isosceles triangle. Can you see the isosceles triangles yet? Great you found it! We have congruent segments CF = FE and our shared leg GF and congruent right angles so by the Side Angle Side Theorem our triangle ∆CGE is an isosceles triangle. We will finally construct our tangent circle to have a radius of BG and a center at G. We know this circle is tangent to our given circles because the radius of Circle G is BG and we know the point B lies on Circle A. This means Circle G touches Circle A in exactly one spot which confirms tangency. We know Circle G is tangent to Circle C for we constructed our isosceles triangle using the segment BE which is the radius of the congruent circle to Circle C. Because these two circles are congruent and constructed as stated above we know the radius of Circle G, GH, connects the two circles. Therefore, H is our point of tangency for Circle G and Circle C. So we have three circles where one of which is tangent to the other two!

Let's now look at the construction of two given disjoint circles.

Our construction is almost identical to the one above, but it is a good review regardless. We will begin this construction with our two given circles. Our first given circle is green with center A and the other given circle is pink with center C. We can see that Circle A has a radius of AB and Circle C has radius of CD. Our next step is to construct a circle with center B congruent to circle C. By marking our center as B and our radius as CD we can construct the blue circle with radius BE. Now we can make a segment CE. This segment will serve as the base of a future isosceles triangle. Let F be the midpoint of CE. Constructing a line through AE and then another line perpendicular to CE through F we get an intersection point G. This intersection point is the vertex of our isosceles triangle. Hence, CG is the final side to our triangle. We know HC = BE, we also know EF = FC, and of course GF is the height for both halves of the triangles, so by the Side Angle Side Theorem ΔEGC is an isosceles triangle. Therefore, our circle with center G and radius BG is tangent to Circle A at the point B and again Circle G is tangent to Circle C at the point H.

Finally, we now can explore what happens to the tangent circle when the two given circles overlap one another.

CLICK HERE TO ANIMATE THE POINT B YOURSELF AND SEE HOW THE CIRCLES INTERACT!

I will not bore my readers by going through the construction of Case 3 for it is basically the same as those above. This construction is merely pulled to one side so that our two given circles overlap. Notice that our tangent circle still maintains a center at G and radius BG. So what is the significance of this exploration if all of the constructions are adjusted versions of one another? For one we can see that our tangent circles encompass different circles and points when the location of our given triangles vary. When Circle A lies on Circle C and inside Circle C our tangent circle lies inside of Circle C, and when Circle A and Circle C are disjoint our tangent circle encompasses Circle A. Is this consistent or does it depend on the base of our isosceles triangle?

As we can see from our model to the left our observations are not consistent. However, we can verify our observation that if we move the positions of our given Circles A and C the position of Circle G will depend on the location of the isosceles triangle, because G is a fixed point on the line AB and is also a vertex of the isosceles triangle. Therefore, when the base CE of the isosceles triangle is lengthened or shortened the vertex G will move along the line AB. Because C is an endpoint for the base it makes sense as Circle C shrinks and enlarges, i.e. its radius lengthens and shortens then the length of CE will change accordingly.

Our tangent circle ceases to exist when our Center C lies on the circle with Center B because then our triangle becomes an equilateral triangle for its sides are the radius of the two congruent circles. This is merely the result of the limit of the circle radius going to zero as the point of tangency is moved to the point of intersection of the two given circles. Below we have a situation when our triangle ceases to exist. This happens when C, E, and G are collinear. But our circles manage to form into what looks like a layered confetti cake! And this is because our radii have aligned and are also collinear. Radius AH meets radius BG which then meets radius GE.

In conclusion, this exploration shows us that it is possible to find a circle tangent to two given circles with arbitrary locations in the graph. Our tangent circle will vary in size according to the position of each given circle. The location of the tangent circle also depends on the location of the given circles for as we have explored above as Circle C changes in length then the base of our isosceles triangle also changes which results in a new position of vertex G/tangent circle with center G on the line AB. The exploration of these circles tangent to one another not only shows us how to construct them, but also shows us that there are different sets of tangent circles within these constructions. There is another set of tangent circles where the tangent circle always lies within the given circles circles, as you play with the animation of our tangent circles try to find the other set of circles tangent to the two given circles, Circle A and Circle C.

Here's a hint to help you get started!

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