Constructing Tangent Circles
In this exploration we consider, and solve, the following task: Construct a circle tangent to two given circles. This problem is not at all obvious to solve and there are three cases we must consider for the first problem: 1) one circle is contained in the interior of the other, 2) the circles intersect one another, and 3) the circles are completely disjoint. We will explore the case for case 1.
Consider case 1. Given two circles, with one contained in the interior of the other, we wish to construct a circle tangent to the given two. We note we say we will construct a circle, since there is not a unique solution to the problem. Call their centers A and B.
What characteristics must our circle have in order to solve our problem? How do we figure out what to do first? Since the direction is completely open to us, let us restrict ourselves slightly by designating a point of tangency on circle A, that is, the point on circle A where our solution circle will be tangent to circle A. Call this point T.
Now that we have chosen T, what next? Notice, if we construct the line defined by the points A and T, then that line, call it m, intersects circle A and forms a right angle between the line and the circle (by right angle between the line and the circle we mean a right angle is formed between our line m and the tangent line of circle A at T).
Why is this important? Since we know our solution circle must be tangent to circle A at point T, and m is the unique line perpendicular to circle A at T, we know the center of our solution circle must lie on the line m. Now, if we can find the center of our circle then we have solved our problem since the raduis would then be the distance from our center to the point T. What can we do to find our center? Observe that the distance between the desired circle and B will be the sum of the radii of circle B and the desired circle since these two circles are tangent. Therefore, if we construct a circle of the same radius as circle B with center at T, we can find our center. Call the point of intersection out of circle A the point D. Let us see the picture.
Since the distance from D to T is the value of the raduis of circle B, the center of our circle is on m between the points A and T. Since the distance from the center of our solution to D is the same as the distance from the desired center to B, then the triangle form by the three points D, B and our desired center is isosceles. This means that the altitude of this triangle with respect to the center of our solution circle intersects the segment DB at the midpoint! Let us construct this segment and call its midpoint M.
Finally, since the atlitude in question intersects the segment DB at M we know the atlitude lies on the perpendicular bisector of the segment DB. From earlier we know the center of the desired circle lies on m, so the the center will the intersection between the perpendicular bisector to the segment DB and m! Call this intersection C.
Let us clean this picture up just a tad.
Now by constructing the circle with center C and radius equal to the distance from C to T, we have our solution!
Remember, we chose the point T where circle C would be tangent to circle A. Since T was arbitrary, this construction works for any point T on the outer circle and there we have infinitely many circles tangent to the two given circles. Here is a link for a script tool of the construction. Click Here! The tool is titled TangentCircleToTwoCircles.
As a last comment we note that the scritp tool appears to work for the other two cases, even though I did not prove it here. If we use the script tool to investigate the other two we get the following convincing evidence that the construction works in all cases.
Also, it is important to note that this is not the only construction. As further exploration for students we could explore tangent circles that look like the following:
That is, given the set-up for case 1, construct a circle tangent to both with the smaller circle contained in the tangent circle. This could be a great way for students to exercise their geometry muscles!