T A N G E N T  C I R C L E

by

Jane Yun

 

Lets begin the investigation with the following problem.

 

 

 

 

Let AD and BC be two circles, and let B 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 AB and then, construct a circle with center B and radius BC.  (The center B of the desired circle BC will lay along a line from the center of the circle A with specified point B.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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

 

 

 

 

 

 

 

 

 

 

 

 

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

 

 

 

 

 

 

 

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 EB, which is from the center E of the desired circle EB to the center B of the second given circle BC.  These segments are always the sum of length of EB and BC(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 EB lies along the perpendicular bisector of the base BC of the isosceles BCE.

 

 

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. 

 

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

 

Given the construction, lets 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 B 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 C 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 C 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 AC, the difference is constant. 

 

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

 

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

 

Then, segment EB = segment CE + segment BC.  By subtracting segment BC, you get segment CE = segment EB – segment BC.

 

Since segment BE = segment CE and segment CE = segment EA + segment AC, segment BE = segment EA + segment AC.

 

By subtracting segment EA, segment BE – segment EA = segment AC, 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 C 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).

 

 

 

 

 

 

 

 

 

 

 

 


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