If you’re reading this blog, you probably know something about reflections in geometry. You have a point (purple in diagram below) and a line (solid). If you reflect the point over the line, you get a new point (orange) on the other side of the line, the same distance as the original point. The segment connecting the two points (dashed) is perpendicular to the line of reflection. (Link to live graph.)

What if you reflect over a circle? I played today with the circle I understand best—the circle of radius 1, centered at the origin.

You still want the line segment to be perpendicular, which is tricky enough. But distance gets messy. Reflecting over a line means taking everything on one side of the line and matching it to something on the other side. Easy to do with two half-planes.

But with a circle? You need everything outside the circle to match up with everything inside the circle. The basic idea is a point outside the circle will match up with a point inside the circle, and that their distances will be reciprocals of each other. If the original point is 2 units from the origin, its reflection will be 1/2 unit from the origin. 3/4 matches with 4/3, 5 with 1/5 and so on.

Whether infinity and zero match up is open to interpretation and not important right now.

Here’s what this looks like.

Go play with the graph. Move the orange point around and start to get a feel for the relationship between the original and its reflection.

Now we’ll do two points and connect them with a straight dotted line segment. Each endpoint is reflected in the circle. (Play with it here.)

What does the reflection of the in-between points look like?

Imagine it. Sketch it. Then go see.

### Like this:

Like Loading...

*Related*

Really cool topic, and explanation of it. Of course….love the use of Desmos!

This is called “inversion in a circle”.

Straight line goes to circle and vv

Straight line touching the circle goes to circle with point on the line and the opposite being the centre of the original circle, and so on…..

This is really cool — nice work!

There’s a whole method in physics for finding the electric potential due to a collection of charges near a conducting surface where you imagine the image of the charges reflected across the surface, then compute the potential of the two sets of charges ignoring the surface. Turns out this gives the correct potential and is much easier than figuring out exactly what happens on the surface of the real conductor.

I assume you are using polar coordinates and R->1/R as the reflection. In complex numbers, the reflection of Z is 1/Z* (the conjugate of 1/Z).

*screeching, sound of car crash in Dan’s head*

You’ll have to walk through that one a little slower for me some time.