
# 10.3: Inversive plane and circlines


Let $$\Omega$$ be a circle with the center $$O$$ and the radius r. Consider the inversion in $$\Omega$$.

Recall that the inverse of $$O$$ is undefined. To deal with this problem it is useful to add to the plane an extra point; it will be called the point at infinity; we will denote it as $$\infty$$. We can assume that ∞ is inverse of $$O$$ and the other way around.

The Euclidean plane with an added point at infinity is called the inversive plane.

We will always assume that any line and half-line contains $$\infty$$.

Recall that circline, means circle or line. Therefore we may say "if a circline contains $$\infty$$, then it is a line" or "a circline that does not contain $$\infty$$ is a circle".

Note that according to Theorem 8.1.1, for any $$\triangle ABC$$ there is a unique circline that passes thru $$A, B$$, and $$C$$ (if $$\triangle ABC$$ is degenerate, then this is a line and if not it is a circle).

Theorem $$\PageIndex{1}$$

In the inversive plane, inverse of a circline is a circline.

Proof

Suppose that $$O$$ denotes the center of the inversion and $$r$$ its radius.

Let $$\Gamma$$ be a circline. Choose three distinct points $$A,B$$, and $$C$$ on $$\Gamma$$. (If $$\triangle ABC$$ is nondegenerate, then $$\Gamma$$ is the circumcircle of $$\triangle ABC$$; if $$\triangle ABC$$ is degenerate, then $$\Gamma$$ is the line passing thru $$A, B$$, and $$C$$.)

Let $$A', B'$$, and $$C'$$ denote the inverses of $$A, B$$, and $$C$$ respectively. Let $$\Gamma'$$ be the circline that passes thru $$A', B'$$, and $$C'$$.

Assume $$D$$ is a point of the inversive plane that is distinct from $$A, C, O$$, and $$\infty$$. Suppose that $$D'$$ denotes the inverse of $$D$$.

By Theorem 10.2.6c, $$D' \in \Gamma'$$ if and only if $$D \in \Gamma$$.

It remains to prove that $$O \in \Gamma \Leftrightarrow \infty \in \Gamma'$$ and $$\infty \in \Gamma \Leftrightarrow O \in \Gamma'$$. Since $$\Gamma$$ is the inverse of $$\Gamma'$$, it is sufficient to prove that

$$\infty \in \Gamma \Leftrightarrow O \in \Gamma'.$$

If $$\infty \in \Gamma$$, then $$\Gamma$$ is a line; or, equivalently, for any $$\varepsilon > 0$$, the circline $$\Gamma$$ contains a point $$P$$ with $$OP > r/\varepsilon$$. For the inversion $$P' \in \Gamma'$$ of $$P$$, we have that $$OP' = r^2/OP < r \cdot \varepsilon$$. That is, the circline $$\Gamma'$$ contains points arbitrarily close to $$O$$. It follows that $$O \in \Gamma'$$. The same way we can check the converse.

Exercise $$\PageIndex{1}$$

Assume that the circle $$\Gamma'$$ is the inverse of the circle $$\Gamma$$. Suppose that $$Q$$ denotes the center of $$\Gamma$$ and $$Q'$$ denotes the inverse of $$Q$$. Show that $$Q'$$ is not the center of $$\Gamma'$$.

Hint

First show that for any $$r>0$$ and for any real numbers $$x,y$$ distinct from 0,

$$\dfrac{r^2}{(x + y)/2} = (\dfrac{r^2}{x} + \dfrac{r^2}{y})/2$$

if and only if $$x = y$$.

Suppose that $$\ell$$ denotes the line passing thru $$Q$$, $$Q'$$, and the center of the inversion $$O$$. Choose an isometry $$\ell \to \mathbb{R}$$ that sends $$O$$ to 0; assume $$x, y \in \mathbb{R}$$ are the values of $$\ell$$ for the two points of intersection $$\ell \cap \Gamma$$; note that $$x \ne y$$. Assume $$r$$ is the radius of the circle of inversion. Then the left hand side above is the coordinate of $$Q'$$ and the right hand side is the coordinate of the center of $$\Gamma'$$.

Assume that a circumtool is a geometric construction tool that produces a circline passing thru any three given points.

Exercise $$\PageIndex{2}$$

Show that with only a circumtool, it is impossible to construct the center of a given circle.

Hint

A solution is given in Section 19.4.

Exercise $$\PageIndex{3}$$

Show that for any pair of tangent circles in the inversive plane, there is an inversion that sends them to a pair of parallel lines.

Hint

Apply an inversion in a circle with the center at the only point of intersection of the circles; then use Theorem $$\PageIndex{2}$$

Theorem $$\PageIndex{2}$$

Consider the inversion of the inversive plane in the circle $$\Omega$$ with the center $$O$$. Then

(a) A line passing thru $$O$$ is inverted into itself.

(b) A line not passing thru $$O$$ is inverted into a circle that passes thru $$O$$, and the other way around.

(c) A circle not passing thru $$O$$ is inverted into a circle not passing thru $$O$$.

Proof

In the proof we use Theorem $$\PageIndex{1}$$ without mentioning it.

(a). Note that if a line passes thru $$O$$, it contains both $$\infty$$ and $$O$$. Therefore, its inverse also contains $$\infty$$ and $$O$$. In particular, the image is a line passing thru $$O$$.

(b). Since any line $$\ell$$ passes thru $$\infty$$, its image $$\ell'$$ has to contain $$O$$. If the line does not contain $$O$$, then $$\ell' \not\ni \infty$$; that is, $$\ell'$$ is not a line. Therefore, $$\ell'$$ is a circle that passes thru $$O$$.

(c). If the circle $$\Gamma$$ does not contain $$O$$, then its image $$\Gamma'$$ does not contain $$\infty$$. Therefore, $$\Gamma'$$ is a circle. Since $$\Gamma \not\ni \infty$$ we get that $$\Gamma' \ni O$$. Hence the result.