Inversion in a Circle

An introduction to circle inversion: the defining relation OP·OP' = k², a ruler-and-compass construction of the inverse point, and what a circle inverts to in all four cases — through the center it becomes a line (general, and the tangent special case), otherwise it stays a circle (general, and the orthogonal circle fixed by the inversion).

Keywords: inversion, circle of inversion, inverse point, tangency preserved, orthogonal circles

Prerequisites: thales-theorem-proof, geometric-mean-proof · Difficulty: advanced

Inversion is defined by a fixed circle of inversion — center $O$, radius $k$. The inverse of a point $P$ (other than $O$) is the point $P^{\prime }$ on ray $OP$ with

A point on the circle is its own inverse; points inside and outside swap. Drag $O$, $K$, or $P$ to explore.

Strategy: construct $P^{\prime }$ from the tangent point $T$, then watch what a circle inverts to across all four cases — through $O$ it becomes a line (general, and the tangent special case), otherwise it stays a circle (general, and the orthogonal circle fixed by the inversion).

The Circle of Inversion

Inversion is defined relative to a fixed circle — center $O$, radius $k$. Here $O$, the point $K$ on the circle, and the point $P$ are all free; drag them to explore.

$OP\cdot O{P}^{\prime }=k^2$

The inverse of $P$ is the point $P^{\prime }$ on ray $OP$ with $OP\cdot O{P}^{\prime }=k^2$. To construct it for an external $P$: the tangent from $P$ touches the circle at $T$, so $\angle OTP=90^{\circ }$ (radius $\perp$ tangent; equivalently $T$ is on the circle with diameter $OP$). Dropping the perpendicular from $T$ to $OP$ at $P^{\prime }$, the altitude of right triangle $OTP$ gives $O{T}^2=OP\cdot O{P}^{\prime }$, so $OP\cdot O{P}^{\prime }=k^2$.

On the Circle, and the Swap

A point on the circle is its own inverse (if $OP=k$ then $O{P}^{\prime }=k$). Points outside map to points inside, and back: here $P$ is outside and $P^{\prime }$ is inside.

A Line Inverts to a Circle Through $O$

A line not through $O$ inverts to a circle through $O$ — and back. Take the tangent line at $K$. For a point $Q$ on it, the inverse $Q^{\prime }$ is the foot of the perpendicular from $K$ to $OQ$: the tangent meets the radius at a right angle ($\angle OKQ=90^{\circ }$), so the altitude of right triangle $OKQ$ gives $O{K}^2=OQ\cdot O{Q}^{\prime }$ — and $OK=k$, so $Q^{\prime }$ is the inverse of $Q$. Moreover $\angle O{Q}^{\prime }K=90^{\circ }$, so by Thales $Q^{\prime }$ lies on the circle with diameter $OK$. As $Q$ runs along the line, $Q^{\prime }$ sweeps that circle, tangent to the circle of inversion at $K$. Conversely, a circle through $O$ inverts to a line — shown in general next.

Any Circle Through $O$ Becomes a Line

The tangent circle was a special case. It can be shown that any circle through $O$ inverts to a line — the same similar-triangle reasoning as the tangent case applies. Here $C$ is built through $O$ and two points $A$, $B$, and its image is the line through their inverses $A^{\prime }$, $B^{\prime }$.

Drag $A$ or $B$ to reshape $C$, and watch its image line follow.

An Orthogonal Circle

Two circles are orthogonal when they cross at right angles — the radius of one at a crossing point is tangent to the other. The circle centered at $P$ through $T$ crosses the circle of inversion at $T$ with $OT\perp PT$, so it is orthogonal to it.

Prove $OS\cdot O{S}^{\prime }=k^2$

Take a point $S$ on the orthogonal circle; the line $OS$ meets it again at $S^{\prime }$. Drop the perpendicular from the center $P$ to $OS$; its foot $M$ is the midpoint of the chord $S{S}^{\prime }$, so $S$ and $S^{\prime }$ are symmetric about $M$:

The right triangles $OMP$ and $SMP$ (right-angled at $M$) give $O{M}^2=O{P}^2-P{M}^2$ and $M{S}^2=P{S}^2-P{M}^2=P{T}^2-P{M}^2$ (as $PS=PT$, both radii), so $O{M}^2-M{S}^2=O{P}^2-P{T}^2$. Finally $△OTP$ is right-angled at $T$ ($OT\perp PT$), so $P{T}^2=O{P}^2-k^2$, and

The Circle Maps to Itself

Since $OS\cdot O{S}^{\prime }=k^2$, the point $S^{\prime }$ is the inverse of $S$ — and it lies on the same circle. This holds for every $S$, so inversion maps the orthogonal circle to itself. Drag $S$ to watch $S^{\prime }$ stay on the circle.

Any Other Circle Becomes a Circle

A circle that does not pass through $O$ inverts to another circle (as with the orthogonal case, we state this rather than reprove it). Three points determine a circle, so $C$ is built through $U$, $V$, $W$, and its image is the circle through their inverses $U^{\prime }$, $V^{\prime }$, $W^{\prime }$.

One caveat: the image circle's center is not the inverse of $C$'s center. Drag $U$, $V$, or $W$ and watch the image circle respond.

Inversion in a circle of radius $k$ sends $P$ to the point $P^{\prime }$ on ray $OP$ with $OP\cdot O{P}^{\prime }=k^2$. Points on the circle are fixed; inside and outside swap. The image of a circle is a line when the circle passes through $O$ (in general, and the tangent-line special case) and a circle otherwise (in general, and the orthogonal circle that maps to itself). Throughout, tangency is preserved.

These are exactly the levers behind the inversion proof of Feuerbach's theorem: invert in a well-chosen circle so the nine-point circle (through the center) becomes a line and the incircle (orthogonal) stays fixed, read off the tangency, and carry it back.

Notes

Constructing $P^{\prime }$ for an external $P$

Draw the tangent from $P$ to the circle, touching at $T$. The radius $OT$ is perpendicular to the tangent, so $\angle OTP=90^{\circ }$ — equivalently, $T$ lies on the circle with diameter $OP$ (Thales). Drop the perpendicular from $T$ to $OP$; its foot is $P^{\prime }$. In the right triangle $OTP$ the altitude $T{P}^{\prime }$ gives the geometric-mean relation

What a circle inverts to — four cases

The image of a circle is a line if the circle passes through $O$, and a circle if it does not:

• Through $O$ $\to$ line. In general, invert two of its points and join them. The special case: the circle on diameter $OK$, tangent to the circle of inversion at $K$, flattens to the tangent line at $K$.
• Not through $O$ $\to$ circle. In general, invert three of its points and take their circumcircle. The special case: a circle orthogonal to the circle of inversion maps to itself (it crosses at right angles — the circle centered at $P$ through $T$ has $OT\perp PT$).

Inversion also preserves tangency, which is what carries the Feuerbach configuration through.