Let's construct a tangent circle.
Case 1: explore tangent circles with one circle is contained within another without touching
1. Start with two circles: one with center at C and a smaller one with center at C' inside the circle C. Circle C has a point D on the the circle. This will be the point of tangency on circle C.
2. Construct a circle with center D and the same radius as circle C'.
3. Construct a line CD. Mark point P where it crosses circle D. Construct line PC'. Find the midpoint of this segment at point M and construct the perpendicular bisector. Where the bisector crosses line CD, mark point N.
4. Construct segment NC'. Circle N with radius ND will be the tangent circle.
If we animate point D and trace point N, we can see that the path of N is an ellipse. For further exploration with this gsp file, click here.
To prove that the path of N is an ellipse, we have to show that the sum of the distance from N to two fixed points is constant. In this example, C and C' are fixed points because they are the center of the two circles. We want to show that the sum of the radii of the two circles (always a constant) is equal to the sum of the distance from the centers of the two circles to the center of the tangent circle: C'R + CD = C'N + CN.
We know that ND = NT = radius of tangent circle.
CD = CN + ND, we can substitute and say that CD = CN + NT.
We also know that C'N = C'T + NT. By substitution, we can say C'T = C'R.
Thus, CN + C'N = CD - NT + C'R + NT. Therefore, CN + C'N = CD + C'R and we have proved that the the path of N is an ellipse.
Case 2: explore with tangent circles with the circles C and C' are disjoint
Using the same projecture, you can create the tangent circle to C and C'.
If we annimate point D and trace point N, we can see that the path of N is a hyperbola. For further exploration with this gsp file, click here.
To prove that the path is a hyperbola, we must show that the difference of the distance from point N to two fixed points is constant. As before, we know that the centers of the two circles are fixed points.
We want to show that C'N - CN = CR + CD.
We know that C'N = CT + NT and ND = CN + CD.
Using that knowledge and substituting CR for CT and NT for ND, we have: C'N - CN = CR +NT - NT + CD = CR +CD. Therefore, we have shown that the path of N is, in fact, a hyperbola with C and C' as the focii.