Definition of inversion in a circle

Question

Let $C$ be a circle with the middle point $O$ and the radius $r$, we say that the points $P$ and $P'$ are inverse points with respect to $C$ if: $$|OP|·|OP'|=r^2$$ Can anyone tell what is the intuition behind this definition? Can this condition be derived?

Answer 1

Inversion in a circle has several nice properties:

• All points on the circle are fixed.
• Every point (except the center of the circle) is mapped onto a point on the same ray out of the center of the circle.
• The regions inside and outside the circle are swapped: the punctured disk consisting of all points inside the circle except the center is mapped onto the entire plane outside the circle, and vice versa.
• Any circle is mapped to a circle or line.
• Any line is mapped to a circle or line.
• The mapping is conformal. If you have a figure in which two curves meet at an angle $\alpha$ at point $P$, the image of the figure will have two curves that meet at an angle $\alpha$ at $P'$, where $P'$ is the image of $P$.

You can construct a circle inversion as follows. Let $O$ be the center of a circle of radius $r$ in the plane $\pi_1$ that passes through a point $R$. Construct a sphere of radius $r$ around $O$, and let $S$ be a point on the sphere along an axis perpendicular to the plane $\pi_1$; that is, $OS = OR = r$ and $\overline{OS} \perp \pi_1.$ The figure below shows a cross-section of this construction using the plane of $\triangle ORS$ as the cross-section plane.

Now suppose you have an arbitrary point $P$ in plane $\pi_1.$ Project that point along the line $PS$ onto the point $Q$ on the sphere. That is, $Q$ is the inverse stereographic projection of $P$ onto the sphere. Reflect the point $Q$ through the plane $\pi_1$ to the point $Q'$, which by symmetry is also on the sphere. Finally, project $Q'$ along the line $Q'S$ onto the point $P'$ on plane $\pi_1.$

In the figure, this process is shown for a point $P$ inside the circle, but the same process that leads from $P$ to $P'$ also leads from $P'$ to $P.$ Moreover, a point on the circle is mapped to itself, so that's one of the "nice properties" listed above.

The points $O,$ $P$, $P'$ are collinear with $O$ not between $P$ and $P'$, so every point (except $O$) is mapped to a point on the same ray out of $O$.

Every point inside the circle (except $O$) is mapped to a point outside, and vice versa.

The remaining properties depend on knowing a few things about the properties of stereographic projection. I will assume this knowledge; you can confirm these here and here or various other places.

The inverse stereographic projection from plane $\pi_1$ to the sphere maps lines and circles to circles on the sphere. The reflection maps circles on the sphere to circles on the sphere to circles on the sphere. The stereographic projection onto plane $\pi_1$ maps circles to lines or circles. Therefore, starting with either a line or a circle, we end up with either a line or a circle.

Finally, all three mappings that we composed in this process are conformal, so the mapping produced by the entire process is conformal.

Now we observe that $\angle QOR = \angle Q'OR,$ from which it follows that $\angle QOS$ and $\angle Q'OS$ are supplementary angles, from which it follows that $\angle QSO$ and $\angle Q'SO$ are complementary angles, from which it follows that the triangles $\triangle OPS$ and $\triangle OSP'$ are similar with $OP : OS = OS : OP',$ and therefore $$ r^2 = (OS)^2 = OP \cdot OP'.$$ So indeed this process produces a circle inversion defined by the given formula.

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Yu might ask if the list of properties at the beginning of the answer uniquely lead to this definition of circle inversion. It does.

If we consider the circle with diameter $\overline{OR},$ it cannot be mapped to another circle, so it must be mapped to a line; that line must pass through $R$ (which is a fixed point under the mapping), in fact it must be mapped to the line through $R$ tangent to the original circle. By projecting the various points of the circle with radius $OR$ along a line through $O$ onto the line through $R,$ we determine the distances $OP$ and $OP'$ for every point $P$ on the circle with diameter $\overline{OR}.$ Moreover, we can rotate this circle and line around the point $O$ in order to find the correspondence between $P$ and $P'$ for any point $P$. We can use this to confirm that we must always have the relationship $OP \cdot OP' = r^2.$

I probably could have derived this formula just from the mapping of the smaller circle to the tangent line without involving stereographic projection, but (1) I really like stereographic projection and (2) the construction with stereographic projections has much more general properties than just mapping that circle to that line.

Answer 2

As I mentioned in my comments, if you take $O$ a the origin, you can describe an inversion about your circle $C$ via the complex transformation $z\longrightarrow \frac{R^2}{\bar{z}}$.

Let $P$ be any point in $\mathbb{C}$ that's distinct from the origin. Define $l(t)$ as the ray $$l(t)=Pt:t\in [0,\infty)$$ Note $P$ is a ray whose initial point is $O$ that emanates in the direction towards $P$. We seek to find $t^*$ so that $$|l(t^*)|\cdot |P|=R^2$$ It's immediate that $t^*=\frac{R^2}{|P|^2}$. Taking $P'=l(t^*)$ gives us $$P'=\frac{PR^2}{|P|^2}$$ which is our prescribed image of $P$ under our inversion with respect to $C$. If we set $P=x+iy$ then $$P'=R^2 \cdot \frac{x+iy}{x^2+y^2}=\frac{R^2}{x-iy}=f(x+iy)$$ where $f(z)=\frac{R^2}{\bar{z}}$.

Answer 3

Important relations for circle inversion with respect to the inverting "mirror" circle with radius equalling $OT =a $ is tangential to circle containing $ P,P'$ at $T$.

$$ OT^2= OP\cdot OP'$$

This arises due to similarity of triangles $ OPT,OT P'$ when considering angles in the alternate segment.

Angles $ OPT,OP'T$ are supplementary.

Also the two circles that get inverted ( not sketched) need not touch the mirroring circle. In this case $( P,P')$ become tangent points.

The segments obey complex relation

$$ z \; z' = a^2 $$

which can be used to derive cartesian coordinates of $(P,P')$.