Given intersecting circles $a$ and $b$, find circle $c$ such that $a$ and $b$ are inverses of each other in $c$

Question

(Just to be clear I'm talking about "inversion" as in inversive geometry)

Given two intersecting circles $a$ and $b$, how do you find circle $c$ such that $a$ inverted about $c$ yields $b$, and $b$ inverted about $c$ yields $a$. Circles $a$ and $b$ may intersect at one point or two points.

To illustrate in the following given the green and blue circles, how do you find the red circle?

Answer 1

From the definition of inversion we can see that each common tangent to the blue and green circle goes through $O$, the center of the red circle. (Namely, each line through $O$ contains the same number of blue points as green points; we get tangency when that number is $1$).

This means that there are two ways to make the green circle into the blue one:

• invert each point separately with respect to the red circle, or
• scale the entire plane by an appropriate linear factor, fixing $O$.

These two ways result in different correspondences between the points on the circles, but the entire blue circle we get in each way is the same.

The value of this is that it is easy to find points on the blue and green circles that correspond to each other under the second transformation:

1. Draw any pair of distinct parallel diameters in the blue and green circles.
2. Connect the upper ends of the diameters with a line.
3. Connect the lower ends of the diameters with a line.
4. The intersection between these two lines must be $O$.
5. The red circle has center $O$ and goes through the points where the green and blue circles intersect.

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This only works when -- like in the diagram in the question -- $O$ is not inside the green/blue circle. If we start out with a situation where this is not the case, and carry out this construction, we'll find a different red circle, but still one that we can invert the blue circle in to get the green or vice versa. In other words, the solution is not unique.

Answer 2

An alternative to my previous answer which handles the non-uniqueness of the reference circle in a more symmetric way:

Notice that at the points where all three circles meet, the red circle bisects the angle between the blue and green curves. (I don't know a geometric proof of this, but it is easy to see for an arbitrary smooth curve that intersects the reference circle, by applying calculus to polar coordinates). This means that the red radius at that point also bisects the supplementary blue-green angle.

Thus:

1. Choose one of the green/blue intersections.
2. Construct the green and blue tangents at that point.
3. Bisect the angles between the green and blue tangents. This gives two bisectors at right angles to each other.
4. Where the bisectors intersect the line connecting the green and blue centers, you find the centers of two different red circles that are both possible solutions.

Answer 3

Just to summarize, these circles actually have a name, "mid-circles". Coxeter discusses them here: "Mid-Circles and Loxodromes".

If the two circles are tangent or non-intersecting there will be one mid-circle. If they intersect at two points there will be two orthogonal mid-circles.

In the case of non-tangent intersection, one mid-circle can be found by finding the circle passing through the two points of intersection with a center located at the intersection of the two circle's mutual tangent lines; the other will be orthogonal to the first at the two intersection points.