T A N G E N T  C I R C L E

by

Jane Yun

LetÕs begin the investigation with the following problem.

Let AD and BC be two circles, and let BÕ be a point on the circle AD.  Construct a circle tangent to the two (given) circles with one point of tangency being the designated point.

Draw a straight ABÕ and then, construct a circle with center BÕ and radius BC.  (The center BÕ of the desired circle BÕCÕ will lay along a line from the center of the circle A with specified point BÕ.

Draw a straight line BCÕ, bisect BCÕ at N, and draw a perpendicular line through N and the segment BCÕ.  Then, let E be the intersection point of the perpendicular line and line ABÕ.

Next, construct a circle with a center E and radius EBÕ.

This is a circle tangent to the two given circles.  Do you see a red circle that is tangent to both two circles?

Now consider the segments EB and EBÕ, which is from the center E of the desired circle EBÕ to the center B of the second given circle BC.  These segments are always the sum of length of EBÕ and BÕCÕ(or BC).  The same distance lies off along the line through the given point from the center of the desired circle.

Therefore, the center E of the tangent circle EBÕ lies along the perpendicular bisector of the base BCÕ of the isosceles ÆBCÕE.

Summary

We have constructed a line through the center of a circle of the same radius as the second of the given circles with the designated point as center.  The intersection of the line and circle will allow construction of the base of the isosceles triangle.  Hence, the intersection of the line and circle will allow location of the center of the desired circle.

LetÕs explore some other problems to set the direction for additional explorations.

Given the construction, letÕs consider the locus of the center of all circles tangent to the two given circles.  By using GSP, we can animate around the circle and trace the locus of the center.

Case 1:  When the second circle is inside the first circle.

By animating point BÕ and tracing the locus of the centers of the tangent circles for this case, I see that the loci of the center of the tangent circles created an ellipse with the foci at the centers of the given circle.  Because the sum of segments, EB & EA, is equal to the sum of radii of two given circles, the sum is a constant. Therefore, the locus of the centers of the tangent circles is an ellipse with foci at the centers of the given circles.

Case 2:  When two given circles intersect.

When CÕ is outside the circle, the locus of the centers of the tangent is an ellipse with foci at the centers of the given circles (by the same reason above).

But when CÕ is inside the circle, the locus of the centers of the tangent created a hyperbola with foci at the centers of the given circles.  Because segment BE - segment EA is equal to the segment ACÕ, the difference is constant.

Notice that the segment ABÕ is the radius of the given circle ABÕ, so the radius is constant.

Then, it was previously shown that BÕCÕE is an isosceles triangle.  It follows that segment BE = segment CÕE (by definition of isosceles triangle).

Then, segment EBÕ = segment CÕE + segment BÕCÕ.  By subtracting segment BÕCÕ, you get segment CÕE = segment EBÕ – segment BÕCÕ.

Since segment BE = segment CÕE and segment CÕE = segment EA + segment ACÕ, segment BE = segment EA + segment ACÕ.

By subtracting segment EA, segment BE – segment EA = segment ACÕ, which is constant.

Therefore, the locus of the centers of the tangent circles is a hyperbola at the centers of the given circles.

Case 3:  When two given circles are disjoint

When CÕ is either inside or outside the circle, the locus of the centers of the tangent is a hyperbola with foci at the centers of the given circles (by the same reason above).