Given two circles and a designated point on one of the circles, construct another circle that is tangent to both circles.
WOW! Can you picture what we are trying to do? We are going to need a great deal of visualization for this investigation. It's a good thing we have Geometer's Sketchpad to help us investigate, demonstrate and explore.
Let's begin with the basic construction. Construct a large circle (dark green) and a smaller circle (red) inside the large one. Select a random point (E) on the larger circle. It will become the point of tangency for the large circle. Now construct a line going through the center of the large circle and this designated point (blue dashed line).
Next, construct a (temporary) circle with the designated point as the center and a radius the same size as the radius of the smaller circle. Draw a segment from the center of the small circle to a point on this temporary circle (brown dashed line). Identify the midpoint of this segment and construct a perpendicular line (red dashed line). Construct a point at the intersection of this perpendicular line and the line drawn from the large circle's center to the designated point (B). This is the center of a circle which is tangent to both circles.
Lastly, construct a circle with this intersection as the center and the desginated point on the arc (light green circle).
Notice what we have, in effect, done. By constructing the perpendicular bisector of the segment connecting the two small circles, we have generated the base of an isosceles triangle. The two congruent sides of this triangle are highlighted in the diagram below. Their length consists of the radius of the small circle plus the radius of the tangent circle. Our construction will maintain these relationships regardless of how we move the circles around.
Next, we will animate point E on the large circle and trace the center point of the tangent circle (B) to see what pattern emerges.
To see the animation, CLICK HERE.
The locus of the center of all circles tangent to the two given circles is an ellipse! An ellipse is defined to be the set of all points such that the sum of the distances between that point and two distinct fixed points (the foci), is a constant. In our case, the center of the light green tangent circle is connected by segments to the centers of the two given circles. The sum of these segments (BA + BC) is the same as the sum of the radii of the two given circles (AD + CE). This sum is a constant, and therefore the locus of the centers of the tangent circles is an ellipse with foci at the centers of the two given circles (A and C).
By altering the placement of the line segment that connects the small and the temporary circle, and therefore changing the intersection point of the perpendicular bisector, we can construct a circle which is tangent to both circles, but surrounds the smaller circle. CLICK HERE to see the construction and its animation. If it is not already set, be sure to mark the intersection (B) for tracing.
Again, the locus of the center of all such tangent circles is an ellipse. And for the same reason as the previous example, the foci are again the centers of the two given circles.
Let's see what we get if we drag the smaller circle so that it is external to the tangent circle. Will a different pattern emerge, or will the same relationships hold? Think about it first, then CLICK HERE to see. Be sure to trace point B.
Surprisingly, the same relationship between the center of the given circles and that of the tangent circle still hold. The sum of the segments connecting the centers of the three circles (AB + BC) still equals the sum of the radii of the two given circles (this time designated by (AD + 2BD) + CE). And the result is still an ellipse with foci at the centers of the two given circles.
One final thought to this investigation. What will the locus of the centers of all tangent circles look like if the two given circles intersect? It seems like a new situation. But if there's any logic to this world, you can surmise that an ellipse will be formed, right? CLICK HERE.
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