## 3.The Definintion of Inversion (AF)(AF')=r^2

1. The inverse of a point inside a circle is a point outside of a circle.
• If (AF) smaller than r, and (AF)(AF')=r^2, then (AF') must be bigger than r, so F' will lie outside of the circle.
2. The inverse of a point outside the circle is a point inside the circle.
• (See above argument)
3. A point on the circle is it's own inverse.
• If (AF)=r, and (AF)(AF')=r^2, then (AF')=r
4. The inverse of the center of the circle is infinity.
• As (AF) approaches 0, (AF') approaches infinity, so that (AF)(AF') can be defined.

## 4. Table 2C

 Shape inverted Inverse of shape Circle through A Line Circle not through A Circle Line through A Line through A Line not through A A Circle

## 5. Constructing the inverse of a circle

1. Construct Circle 1 (c1) with A as it's center.
2. Construct Circle 2 (c2) with center B, and Radius AB.
3. Construct a Point P on c2.
4. Construct Ray AP.
5. Label Point R at the intersection of RayAP and c1
6. Construct Segment AP
7. Choose a point on c1, label it T. Make segment AT.
8. Select AP, and then AT, and mark the ratio.
9. Choose A as your Center of Dilation, and dilate P by the marked ratio above.
10. Label this point P'.
11. Trace P'
12. Move P around c2 to get the inverse of c2.

## 6. The inverse of a line not through the center of the circle of inversion.

Script
1. Construct c1 with A as center.
2. Construct line j, not intersecting c1.
3. Choose P on j, to be inverted.
4. Construct Ray AP
5. Mark R as the intersection of AP and c1.
6. Construct Segment AP
7. Make a radius segment from A to a point on c1.
8. Select radius segment, hold down Shift key and select Segment AP. Mark ratio.
9. Select A to be your Center for Dilation.
10. Dilate R by marked ratio.
11. Name dilated point P', trace this point.
12. Move P along the line to get the inverse of line j.

## 7. P and Q are related by inversion.

The definition of inversion through circle of radius r centered at A: The inverse F' of a point F is such that A,F, and F' are collinear, and (AF)(AF')=r^2. Using the sketch above, a line can be drawn through T,P, and Q. (They are collinear.) Calculations show that (TP)(TQ)=(TU)^2. Thus, P is the inverse of Q, TU is the radius, and T is the center of the circle of inversion.