Fix a circle with center *O* and radius *OA = r.* For each point *P* other than the
center *O,* the point *inverse* to *P* in the circle *O* is that point *P'* on the line *OP*
(extended in the direction of *P* if needed) so that the proportion

is satisfied. An equivalent condition is that

In order to cover the special case that occurs when *P* is the center
*O* of the circle, we'll add a point to the plane, and let *O* and be inverses of each other.

We'll use the notation *P**C* to indicate the inverse point *P'.* Read "*P**C*" as "*P* through
*C.*" More generally, when *B* is any plane geometric figure (point, line, circle, etc.) and *C* is a circle,
we'll use the notation *B**C* for the result of inverting *B* in the circle *C.*
We'll also use this notation for reflecting across a straight line. So, if *B* is any plain geometric figure, and *L*
is a straight line, then *B**L* is the figure that results from reflecting *B* across
*L.*

Note that inversion in a circle is an involution. That is, application of inversion in the same circle a second
time undoes the first application. Otherwise said, it is an operation that when done twice yields the identity operation.
Algebraically, this observation is the identity (*B**C*)*C* =
*B,*
where *B* is any plane figure.

Also note that a point *P* is its own inverse, that is, *P**C* = *P,* if and only if it lies on the circumference of the circle. In particular, *C**C* = *C.*

The standard construction to invert a point in a circle uses both compass and
straightedge, so we won't be able to use it. Our construction using just a compass will come later. Still, the standard
construction is interesting. We only need to consider points *P* which either lie outside the circle, or lie inside the
circle but are not the center of the circle; points on the circle are their own inverses, and the center has as its inverse.

Let there be a circle with center *O* and radius
*OA = r,* and let *P* be a point. To invert *P* in the circle, first draw the line *OP.* Now, if
*P* happens to be a point outside the circle, then draw the tangents *PB* and *PC* to the circle, then draw the
line *BC.* Where these two lines meet will be the inverse point *P'.* Since the two right triangles *OBP* and
*BP'P* are similar, therefore the proportion *OP / r = r / OP'* holds.

But if *P* happens to be a point inside the circle (except the center),
then draw the line *BC* perpendicular to *OP,* and let *B* and *C* be the points of intersection of that line
with the circle. Then draw tangents to the circle at *B* and *C.* In this case, the inverse point *P'* is the
intersection of the two tangent lines. Note again the two similar right triangles, from which it follows that the identical
proportion holds.

A third case occurs when *P* lies on the circle, and in that case let the inverse point *P'* be
the same as *P* itself.

The fourth case occurs when *P* is the center *O* of the circle, and as mentioned
above, *O* and are inverses.

We will
invert lines and circles in our circle with center *O* and radius *OA* = *r.*

First, consider a line *BC* which doesn't pass through the
center *O.* We'll show that its inverse is a circle passing through the center of the circle. Let *D* be the foot of
the perpendicular drawn from *O* to the line *BC.* Let *D'* be the point inverse to *D.* Draw the circle
with diameter *OD'.* Let *E* be an arbitrary point on the line *BC,* and let *E'*
be where the ray *OE*
meets the circle with diameter *OD'.* Then triangle *OE'D'* is a right triangle, and it's similar to the right
triangle *ODE.* Hence, we have the proportion

Since inversion is an involution, it is clear that any circle passing through the center of our circle will invert to a straight line.

Should a
line *BC* pass through the center *O,* then it is clear that its inverse will be itself.

We'll start with the case where *O* lies outside the circle *BCD.*

Let *E* be any point on the circle
*BCD,* and *F* the other point where the line *OE* intersects the circle *BCD.* Let *OT* be a line from
*O* tangent to the circle *BCD* at the point *T,*
and let *t* denote the length *OT.*
Then by Euclid's
Proposition III.35, *OE OF = t*^{2}.

We're looking for a point *E'* on the line *OE*
such that *OE OE' = r*^{2}. Dividing this equation by the equation
*OE OF = t*^{2}, we recognize that what we need is a point *E'* so that

Therefore, *OE'* is a
constant times *OF.* This means that when we apply a scaling to the plane by a factor of
*t*^{2} */ r*^{2} with the fixed point of the plane being *O,* then points on the circle
*BCD* will be inverted to points in the scaled circle. Note that both the original and scaled circles have the same tangent
lines from the point *O.*

The other case occurs when *O* lies inside the circle *BCD.* It may seem that
because there are no tangents from *O* to the circle *BCD* we've run into serious problems. But we didn't actually use
the tangent lines in the first case; all we needed to know was that the product *OE OF* was a constant. In this case,
it's also a constant.

As
before, let *E* be any point on the circle *BCD,* and *F* the other point where the line *OE* intersects the
circle *BCD.* (This time, *F* is on the other side of *O.*)
Then by Euclid's Proposition III.36, the product
*OE OF* is a constant, call it *k.*

We're still looking for a point *E'* on the line *OE* such that
*OE OE' = r*^{2}. Dividing this equation by the equation *OE OF = k*, we
recognize that what we need is a point *E'* so that

Therefore, *OE'* is a constant times
*OF.* This means that when we apply a scaling to the plane by a factor of *k / r*^{2} with the fixed
point of the plane being *O,* then points on the circle *BCD* will be inverted to points in the scaled circle.

We conclude that when one circle is inverted in another, and the center of the other is not on the circumference of the first, then the result is a circle. But if the center of the other is on the circumfere nce of the first, then the result is a straight line.

Now examine what happens to
the inverted point *D = BOT* of a point *B* in the circle *OT* as the circle
enlarges. In the diagram, as you drag point *A* toward *S,* the circle *OT* approaces the line *ST* and the
inverse point *D* approaches *C = BST,* the reflection of *C* in the line
*ST.* Thus, the operation of reflection in a line is a limit of inversions in circles.

Next part: 4. Stereographic projection

March, 2002.

David E. Joyce

Department of Mathematics and Computer
Science

Clark University

Worcester, MA 01610

Email: djoyce@clarku.edu

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