Given two circles and a point on one of the circles.
Construct a circle tangent to the two circles with one point of tangency being the designated point.
In this problem, we have two circles, AB and CD, plus an arbitary point P on one of the circles. We are asked to find a circle tangent to both green circles, with P one of the points of tangency.
This is a fairly complex construction. The first thing we need to do is copy CD, the other circle, with a center at P. I draw that orange circle. Also, we'll construct a line through A and P which we'll need later.
There's something interesting here. Any point on the orange circle is the same distance from P as center C is from its circle. We know our tangency points will be P and somewhere on circle CD, so the orange circle will be important. For symmetry-related reasons, the key points will be along line AP. We'll mark the two intersections as O (outside in this graph) and I (inside here).
Let's take the segment OC, find its midpoint, and construct its perpendicular bisector. That will intersect line AE at some point, which we'll label X. If we connect C and X, we've achieved congruent triangles XHO (shown in purple) and XHC (shown in blue). They're both right triangles from the perpendicular part, segment XH appears in both triangles, and length HO = HC from the bisector part. Side-Angle-Side makes congruence.
Because circles PO and CD are the same size, the distance from O to P is the same as the distance from C along segment XC to the edge of the circle. We'll mark that intersection as point F. Since length XO = XC and PO = FC, XP = XF. A circle with center at X and radius XP (or XF) will touch both those points. Because F is the closest point to X, the remainder of circle CD will stay outside this tangent circle. Because point P is inside point O, the remainder of circle AB will stay inside the tangent circle. The light blue double tangent circle is what we want.
What about using point I? Let's do the same thing by connecting I and C, drawing a perpendicular bisector, and finding the bisector's intersection with AP. We'll call that point Y. As before, we form congruent triangles, YGI and YGC.
Unlike the prior time, we don't want to look inside from C towards Y, because P is outside I. We want to extend segment YC and utilize the other intersection point E. In this case, both circle AB and circle CD will be inside the tangent circle centered at Y with radius YP. I drew that circle in crimson red; it runs off the screen a little bit.
For this image, I hid the construction lines so we can see both tangent circles, light blue centered at X, and crimson red centered at Y.
The construction we built will cover many possible cases. For instance, it works when AB and CD are completely separate, when they're tangent, when they overlap, and when one circle lies inside the other. The only trick is to watch for inside and outside. In the example we built, blue F was between X and C, while red E was past C. Sometimes the location of tangency flips and becomes the other intersection of line and circle CD. For instance, in the following schematic tangent points are opposite both E and F.
If you have Geometer's Sketchpad, you might want the TangentCirclesRAM.gsp sketch file.
I also created an applet using Java Sketchpad. You can move A, B, C, D, and P to watch changes to the blue and red circles (though not the construction details). The applet is available on the TangentCircles.html page.