# Wedging an Angle between Two Circles Produces a Limaçon

Initializing live version

Requires a Wolfram Notebook System

Interact on desktop, mobile and cloud with the free Wolfram Player or other Wolfram Language products.

Consider two nonintersecting circles a fixed distance apart. Let be one of the points of perpendicular intersection of the tangents to the two circles. As moves subject to that constraint, what curve does it trace out?

[more]

The surprising answer is that even if the two circles are different sizes, or even if we replace the right angle with another wedge angle, the curve always traces out a limaçon.

[less]

Contributed by: William Paulsen (August 2022)
Open content licensed under CC BY-NC-SA

## Details

There are some special cases with well-known results. If the radii of both circles are reduced to 0, so that the angle is wedged between two points, sweeps out part of a circle, since these are the points where the subtended angle of the line segment between the points is constant. The angle would then be an inscribed angle of the circle [1, p. 283]. A circle is, in fact, a special case of a limaçon.

If only one of the radii is reduced to 0, so that there is a point and a circle, and the angle is , this gives the definition of the pedal curve of the circle. In general, the pedal curve of a plane curve and given fixed pedal point is the locus of all orthogonal projections of the point onto the tangent lines of the curve . The pedal curve of a circle is known to be a limaçon [2, p. 163].

The centers of the two circles are placed units to the left and right of the origin. (In the figure, is always 2.) Denote the radii of the circles by and . Let be the angle of the wedge (the two green lines) between the circles. The parameter is introduced so that the tangent point on the left circle is at an angle of . Using simple geometry, the standard angle for the tangent point on the right circle is . When and , the figure becomes symmetric, so the intersection point is on the axis.

The coordinates of the point of tangency for the left circle are

,

and the slope of the tangent line is

.

Thus, the left green tangent line can be expressed as

.

Likewise, the tangent point on the right circle is

,

so the right green tangent line can be expressed by

.

These green lines intersect at the point

.

Note that can never be 0 or , so the denominator is never zero. Thus, this can be scaled by to remove the denominators. Also, can be shifted by the constant . Finally, and can be found such that

and

.

Then the parametric equations become

,

.

It is already starting to look like the equation of a limaçon. To finish, rotate the coordinate system by letting

,

.

Then,

,

.

Finally, change the parameter by letting , to produce the equations

,

,

which is the equation of a limaçon [3, p. 113].

References

[1] D. C. Alexander and G. M. Koeberlein, Elementary Geometry for College Students, 5th ed., Belmont, CA: Brooks/Cole, 2011.

[2] J. Edwards, An Elementary Treatise on the Differential Calculus, 2nd ed., London: Macmillan and Co., 1892.

[3] J. D. Lawrence, A Catalog of Special Plane Curves, New York: Dover Publications, 1972.

## Permanent Citation

William Paulsen

 Feedback (field required) Email (field required) Name Occupation Organization Note: Your message & contact information may be shared with the author of any specific Demonstration for which you give feedback. Send