**Q:** ** **How
does a geometer capture a lion in the desert?

**A:** Build a circular cage in the desert, enter
it, and lock it. Now perform an inversion with respect to the cage. Then you
are outside and the lion is locked in the cage.

a mathematical joke from before 1938

**Definitions.** An ** inversion
with respect to a circle** Γ is a transformation from the extended plane (the plane with ∞, the point at infinity, added) to
itself that takes

Note that an inversion takes the inside of the circle to the outside and vice versa and that the inversion takes any line through the center to itself. Because of this, inversion can be thought of as a reflection in the circle.

We strongly suggest that the reader play with inversions by
using dynamic geometry software such as *Geometers
Sketchpad*^{®}, *Cabri*^{®},
or *Cinderella*^{®}. Using any
of these you may construct the image, *P*′,* *of*
P *under the inversion through the circle*
*Γ* *as
follows:

* *

*If
P is inside *Γ *then
draw through P the line perpendicular to the ray CP. Let S and R be the intersections
of this line with *Γ*. Then P*′*
is the intersection of the lines tangent to *Γ* at S and R. *(To construct the tangents,
note that lines tangent to a circle are perpendicular to the radius of the
circle.)

* *

*If
P is outside *Γ* then
draw the two tangent lines from P to *Γ*.
Let S and R be the points of tangency on *Γ*.
Then P*′* is the intersection of the line SR with the
ray CP. *(The points *S* and *R* are the intersections of Γ with the circle with diameter *CP*. *Why*?)

Click here for dynamic JAVA
applets using *Cinderella*^{®}
to implement the constructions and showing the images of

circles and lines under inversions. We suggest that the reader explore these applets.

**a. ***Prove that these constructions do construct
inversive pairs.*

The purpose of this part is to explore and better understand inversion, but it will not be directly used in the other parts of this problem.

**b. ***Show that an inversion takes each circle
orthogonal to the circle of inversion to itself.*

Two ** circles are orthogonal** if, at each
point of intersection, the angle between the tangent lines is 90°. (Note that,
at these points, the radius of one circle is tangent to the other circle.)

**c. ***Show that an inversion takes a circle
through the center of inversion to a line not through the center, and vice
versa.* *What happens in the special
cases when either the circle or the straight line intersects the circle of
inversion*? Note that the line is parallel to the line tangent to the circle
at *C*.

Look at Figure 7 and prove that
Δ*CPQ *and Δ*CQ*′*P*′ are similar triangles.

**d.*** An inversion takes circles not through the
center of inversion to circles not through the center. *Note: The
circumference of a circle inverts to another circle but the centers of these
circles are on the same ray from *C*
though **not** an inversive pair.

Look at Figure 8. If *P*,
*Q*, *X* invert to *P*′, *Q*′, *X*′, then show
that

*P*′ *X*′ *Q*′ = *P* *X* *Q*
= right angle,

by looking for similar triangles. Thus, argue that as *X* varies around the circle with diameter
*PQ* then *X*′ varies
around the circle with diameter *Q*′*P*′.

Go to:

Problem 1. Circles in the Plane, which explores some geometry of circles.

Problem 2. Inversions in Circles, which explores properties of inversion in circles.

Problem 3. Applications of Inversions to the Peaucellier-Lipkin Linkage, which explores the applications of Problems 1 and 2 to an understanding of the linkage.