Investigation of the inverse
Location of the inverse of a point inside a circle:
AR (the radius of the circle of inversion) = 1 where A is the center
of the circle and R is on the circle and a point F lies on the ray
AR between the center of the circle A and point R . Because F is inside
the circle, AF is less than 1, but greater than zero. (AF') = (AR)(AR)
/ (AF) so (AF') = (1) / (a number less than 1, but greater than zero) therefore,
(AF') is greater than 1, which puts the point F' ( the inverse of F) outside
of the circle
The inverse of a point outside the circle is inside the circle:
Let AR (the radius of the circle of inversion) = 1 where A is the
center of the circle and R is on the circle. Becuase F is outside the circle,
AF is greater than 1.
(AF') = (AR)(AR) / (AF) so (AF') = (1) / (a number greater than 1)
therefore, (AF') is less than 1, which puts the point F' ( the inverse
of F) inside of the circle
Are there points that are their own inverses?
Yes, points that lie on the circle of inversion are there own inverses.
Let point F be on the circle of inversion where AR (the radius of the circle
of inversion) = 1
with A is the center of the circle and R is on the circle therefore
AF = AR = 1 and (AF') = (AR)(AR) / (AF) so (AF') = (1) / (1) = 1 which
meansF' ( the inverse of F) must also lie on the circle of inversion.
Where is the inverse of the center of the circle?
The inverse of the center of the circle is at infinity. Let Point F be
at point A ( the center of the circle of inversion) and AR (the radius
of the circle of inversion) = 1
where A is the center of the circle and R is on the circle. Becuase
F is on the center of the circle AF = 0. (AF') = (AR)(AR) / (AF) so (AF')
= (1) / (0) = infinity
therefore, F' ( the inverse of F) is at infinity.