Inversion transform with center (or pole) $C$ and power $k^2$ is defined by:

$$\tag{1}J_{C,k}:M \leftrightarrow M' \ \ \ \ \ \iff \ \ \ \ \ \ \ \vec{CM'}=\frac{k^2}{||\vec{CM}||^2} \ \vec{CM} $$

It is an "involutive" transform: $M'$ is the image of $M$ iff $M$ is the image of $M'$. This explains the double arrow.

This transform, credited to Magnus and more or less at the same time to Plücker in the early 1830s, is detailed in many books/sites.

See Appendix 1 for a comment about inversion properties displayed on Fig. 1.

enter image description here

Fig. 1. : Inversion with center $C$ and power 1. Shapes in black are exchanged by inversion with shapes in red. The blue circle, called "circle of inversion", is the locus of invariant points.

But there are ill-known features that I would like to gather.

Moreover, I would like to enlarge this "quest" to interesting (not fully standard) questions/applications of inversion.

For the initialization of this "collection", I propose 3 themes and a compendium of interesting facts:

  1. "Champagne glass inversion" : Fig. 2 shows how inversion with center $C$ and radius of inversion $k$ (we find back the large and the small fish, inverted images one of the other...) can be realized by a 3 steps $P\Gamma P^{-1}$ operation with respect to a paraboloid $\Pi$, where:
  • $P$ is the vertical projection from $\Pi$ onto the horizontal plane, and

  • $\Gamma$ is the conical projection from $\Pi$ to $\Pi$ with center $C_0(a,b,a^2+b^2-k^2)$ where $(a,b)$ are the coordinates of $C$.

This non-classical way to describe inversion deserves an explanation that we have placed in Appendix 2.

enter image description here

Fig. 2. Inversion by a conical projection $\Pi \to \Pi$.

  1. "Bubble inversion" This is a cousin representation of the first one with a sphere instead of a parabola (see Fig. 3). It necessitates to use 2 steps with the two stereographic projections $S_N$ and $S_S$ from the North and South pole resp. with respect to the equatorial plane. Take a look at Fig. 3.

Consider a certain figure, say the circle in red. It is transformed by $S_N$ into the circle on the unit sphere materialized by black little diamonds ; this circle is transformed in turn by $S_S$ into the blue circle on the equatorial plane... which is the inversive image of the initial red circle. Briefly said :

$$\tag{$\star$}I=S_S \circ S_N$$

This shouldn't come as a surprise because stereographic projections are themselves 3D inversions.

See (composition of stereographic projections is inversion through the ball - a geometric way) for a proof.

enter image description here

Fig. 3. (planar) inversion realized by combining two stereographic projections (formula ($\star$)).

  1. "How inversion transform is connected with linear algebra" (in fact connected with item 1) ; this issue looks paradoxical because inversion is definitely not a linear transform. In fact, there exists a group, the anallagmatic group, with a $4 \times 4$ linear representation (also called conformal geometry group). Moreover, this group includes another category of non-linear transforms, translations. Reference: the very good book "Riemannian Geometry" by S. Gallot, D. Hullin, J. Lafontaine, 2nd edition 1993. Universitext, Springer, pages 175-176. Here is how this correspondence is done (Explanations will be found in this book or in the very interesting development by @MvG cited in 4a):

$$\text{If} \ J \ \text{is the basic inversion (center 0, power 1):}$$ $$[J]:=\begin{pmatrix}0&0&0&1\\0&1&0&0\\0&0&1&0\\1&0&0&0\end{pmatrix}.$$ $$\text{If} \ H_r \ \text{is the homothety with ratio} \ r \ \text{and center} \ 0: $$ $$[H_r]:=\begin{pmatrix}1&0&0&0\\0&r&0&0\\0&0&r&0\\0&0&0&r^2\end{pmatrix}.$$

$$\text{If} \ T_V \ \text{is the translation by vector} \ V=\binom{a}{b} :$$ $$[T_V]:=\begin{pmatrix}1&0&0&0\\2a&1&0&0\\2b&0&1&0\\a^2+b^2&a&b&1\end{pmatrix}.$$

All these transformations preserve the following quadratic form:


where $x_0$ should be considered as a homogeneous coordinate (i.e., is there to take into account the projective dimension). The signature of $q$ being $(+++-)$, we are dealing with elements of the group classically denoted $O(3,1)$.

This correspondence with 2D transforms can be extended in a straightforward way to higher dimensions (for $n$D transforms, the correspondence is with $(n+2) \times (n+2)$ matrices).

  1. This reference (jump to slides beginning at slide 130) uses linear algebra in $SL(2,\mathbb{C})$.

  2. (specific to the 2D case): Complex representation :

Inversion with the unit circle $S^1=\{z\in\mathbb{C}\,|\,|z|=1$} as invariant circle is the map

$$I : \begin{cases}\ \begin{array}{ccc}\mathbb{C}\setminus\{0\}&\longrightarrow&\mathbb{C}\setminus\{0\}\\z&\mapsto&\frac{1}{\overline z}\end{array}\end{cases}$$

Result : Any (2D) rotation can be obtained as a certain combination of translations and inversions.

It is a consequence of the following identity (as given in https://mathoverflow.net/q/19965) :

$$e^{i \theta} + \frac{1}{-e^{-i \theta} + \frac{1}{e^{i \theta} + \frac{1}{z}}} = - e^{2 i \theta} z. \tag{%}$$

If we conjugate both sides of (%), we get :

$$e^{-i \theta} + \frac{1}{\overline{-e^{-i \theta} + \frac{1}{\overline{e^{-i \theta} + \frac{1}{\overline{z}}}}}} = - e^{-2 i \theta}\overline{z}. \tag{%%}$$

(The LHS of (%%) can be written in a symbolic way as : $T_{e^{-i \theta}}\circ I \circ T_{-e^{-i \theta}} \circ I \circ T_{e^{-i \theta}} \circ I$).

If we take $\theta=0$ in (%%), one gets :

$$1+ \frac{1}{\overline{-1 + \frac{1}{\overline{1 + \frac{1}{\overline{z}}}}}} = - \overline{z}. \tag{%%%}$$

(Please note the conjugation bar).

  1. Preservation of crossratio:


(see figure 4).

enter image description here

Fig. 4: Preservation of crossratio.

This property (preservation of cross ratio) is shared with projective transforms.

(take care: the image of line segment $AB$ isn't line segment $A'B'$ for example). For a proof, see the very didactic presentation here where crossratio is used for defining hyperbolic geometry distance.

  1. A compendium of interesting documents/issues

My (enlarged) question is : can you give examples (of your own) of interesting, possibly non-usual, uses of inversion ?

Appendix 1: Recall of properties of inversion as depicted by Fig. 1:

  • The image of a straight line is in general a circle passing through the pole $C$ ; exceptionally, when the straight line passes through the origin, its image is the line itself (caution: in the latter case, the straight line is "globally" invariant, but its points, in general, aren't invariant).

  • The image $\Gamma'$ of a circle $\Gamma$ not passing through the origin is a circle of the same type (caution: the center of $\Gamma'$ is not the image of the center of $\Gamma$ ; nevertheless these centers are aligned with the pole of inversion). The image of a circle passing through the origin is a straight line, as said before.

  • The image of the black fish made of line segments is the red fish made of circular arcs swimming in its blue round fishbowl.

Appendix 2: Explanation of inversion depicted in Fig. 2.

Let us begin by another definition of inversion.

$$\tag{3}J_{C,k}:M \to M' \ \ \ \ \iff \ \ \ \ \ \exists \vec{U} \ \text{(unit norm vector) s.t.} \ \begin{cases}\vec{CM}=\lambda \vec{U}\\ \vec{CM'}=\lambda' \vec{U}\end{cases} \ \text{and} \ \lambda\lambda'=k^2$$

Let $M(x,y)$ and $M'(x',y')$ exchanged by inversion $J_{C,k}$.

Let $M_0(x,y,x^2+y^2)$ and $M_0'(x',y',x'^2+y'^2)$ be their "lifted version" on paraboloid $\Pi$.

Let $N(X,Y,Z) $ be any point of line $C_0M_0$. We can define an abscissa $\mu$ on this line for point $N$ in the following way:

$$\tag{$\star$}\vec{C_0N}=\mu \vec{C_0M_0} \ \ \ \ \iff \ \ \ \ \begin{cases}X=a+\mu(x-a)\\ Y=b+\mu(y-b)\\ Z=c+\mu(x^2+y^2-c)\end{cases}$$

with $$c:=a^2+b^2-k^2.$$

Please note that for $\mu=1$, $N$ is in $M_0$. Let $\lambda'$ be the value of $\mu$ associated with $M'_0$.

Thus $N \in \Pi$ in two cases: when $\mu=1$ ($N=M_0$) and when $\mu=\lambda'$ ($N=M'_0$). As

$$\tag{3}N \in \Pi \ \iff \ Z=X^2+Y^2,$$

plugging in (3) the expressions of $X,Y,Z$ in $(\star)$, and replacing $c$ by its defining expression, we obtain the following quadratic equation with unknown $\mu$:

$$\mu^2[(x-a)^2+(y-b)^2] \ + \ \mu[\cdots] \ + \ k^2=0$$

Using the classical formula for the roots' product, we have

$$\lambda' \times 1 =\dfrac{k^2}{(x-a)^2+(y-b)^2}=\dfrac{k^2}{\|\vec{CM}\|^2}$$

We find back here the definition of inversion given in (2) (recall that $\vec{CM}$ hasn't been normalized).

  • $\begingroup$ See the answer here about the (implicit) use of inversion under the form $x \to x/\|x\|^2$ to prove that $d(z,w):=\frac{|z-w|}{\sqrt{1+|z|^2}\sqrt{1+|w|^2}}$ is a distance. See as well here. $\endgroup$
    – Jean Marie
    Oct 21, 2021 at 8:15
  • 1
    $\begingroup$ In the same vein, one can deduce Ptolemy's inequality or identity from triangle inequality (or equality) using inversion as established here. $\endgroup$
    – Jean Marie
    Nov 10, 2021 at 13:17
  • $\begingroup$ For a detailed history, see Patterson, The Origins of the Geometric Principle of Inversion $\endgroup$
    – brainjam
    Nov 17, 2021 at 4:22
  • $\begingroup$ @brainjam Thank you very much. I will try to find this article without a paywall... $\endgroup$
    – Jean Marie
    Nov 17, 2021 at 9:01
  • 1
    $\begingroup$ Isodynamic point $X_{15}$ has many interesting properties, in particular using inversion. $\endgroup$
    – Jean Marie
    Aug 3, 2023 at 7:35

8 Answers 8


An application of the inversion to mechanical linkages (converting circular to linear motion): Peaucellier–Lipkin linkage.

Sylvester writes that when he showed a model to (Lord) Kelvin, he “nursed it as if it had been his own child, and when a motion was made to relieve him of it, replied ‘No! I have not had nearly enough of it—it is the most beautiful thing I have ever seen in my life.’”

  • 1
    $\begingroup$ Thank you for reminding me this beautiful, and ill known indeed, way of doing inversion (and the funny quote). $\endgroup$
    – Jean Marie
    Jan 29, 2018 at 21:13

Chapter 8 of Evan Chen's book 'Euclidean Geometry in Mathematical Olympiads' has a pretty neat application of inversion to the Shoemaker's Knife Problem. For more information see - http://www.math.ubc.ca/~cass/courses/m308-03b/projects-03b/hunter/hunter.html


In projective geometry, given four collinear points $A,B,C,D$, the cross product $(A,B;C,D)$ is defined as $\dfrac{AC\cdot DB}{CB\cdot AD}$.

If $(A,B;C,D)=-1$, the points are said to be harmonic, and $D$ is said to be the harmonic conjugate of $C$ with respect to $A,B$. More generally, the pairs of points $(A,B)$ and $(C,D)$ are said to be harmonic conjugates because $A$ is also the harmonic conjugate of $B$ with respect to $(C,D).$

In inversive geometry, let $(AB)$ denote the circle with diameter $AB$. For a point $C$ collinear with $A,B$, let $D$ be the inverse of $C$ with respect to $(AB)$. Then it's not difficult to show that the set $(A,B,C,D)$ is harmonic. The pair $(C,D)$ inverts to $(D,C)$, and the circle $(CD)$ inverts to itself. Similarly the circle $(AB)$ inverts to itself with respect to $(CD)$

Referring to the concept of $n-$spheres, a circle is a $1-$sphere and a point pair is a $0-$sphere. From this perspective we can say that harmonic conjugation for $0-$spheres and inversion for $1-$spheres are very much the same thing. Awareness of this can be helpful in working with one or the other. For example, the concepts of projective involutions and coaxal pencils of circles are related in this light. As are the idea of harmonic conjugate point pairs and orthogonal circles.

This correspondence is "ill known" only in the sense that projective geometry and inversive geometry tend to be explained as separate topics, each with their own frameworks and tools. And even when a student has been exposed to both topics, the correspondence may not be obvious until it is pointed out.

  • $\begingroup$ You fill indeed a gap by recalling this ill-known (especially now) correspondence between harmonic division and inversion. Thanks! $\endgroup$
    – Jean Marie
    Jan 7, 2022 at 16:05

Perhaps this is not what you are interested in, but here it goes: the inversion with respect to the circle $S^1=\{z\in\mathbb{C}\,|\,|z|=1$} is simply the map$$\begin{array}{ccc}\mathbb{C}\setminus\{0\}&\longrightarrow&\mathbb{C}\setminus\{0\}\\z&\mapsto&\overline z^{\,-1}.\end{array}$$

  • $\begingroup$ It's in the scope of my question : aspects of inversion that don't come immediately to mind when you see the definition... $\endgroup$
    – Jean Marie
    Jan 29, 2018 at 11:55

Several properties of the circles in a Pappus chain can be derived using circle inversion, as this picture borrowed from wikipedia shows:

enter image description here

Along the same idea, a construction based on inversion has been proposed for drawing the figure with TikZ in an answer to The Pappus Chain on tex.stackexchange.

The Pappus chain is also related to Steiner chains, which are in turn related to Soddy's hexlet and the Dupin cyclides mentioned by the OP.


In your question you discuss inversion via two quadrics, i.e. the paraboloid and the sphere.

The book Pedoe, Geometry: A Comprehensive Course has an excellent description of this phenomenon. The cases of paraboloid and sphere are essentially the same, tied together by a projective transformation. I'll refer the reader, if interested, to the book for the details (pg 151-152).

enter image description here

The illustration above is taken from the book. The essential idea is that we are using stereoscopic projection for a paraboloid $\Pi$ instead of a sphere. The projection point corresponding to the north pole $N$ of the sphere is the point at infinity in the direction of the $Z$ axis. The circle is lifted to an ellipse that is a planar section of $\Pi$, and the point $P$ is the pole of that plane with respect to $\Pi$.

The point $A$ is lifted to a point $Q$ on $\Pi$. The line $PQ$ intersects $\Pi$ in a second point $Q'$, which projects down to $A'$, the inverse of $A$.

This is simply a rewrite of the champagne glass inversion in the question, but translates it into a more projective form - incidence, projection, poles, polars. It can be translated word for word to work for the sphere $\Sigma$ and arbitrary circles, not just the equatorial circle.

One immediate application of this approach is to find the point $O'$ on the sphere $\Sigma$ that projects to the center $O$ of a circle. Since we know that $O$ inverts wrt the circle to the point at infinity, we can conclude that $O'$ is the intersection of the line $NP$ and $\Sigma$. And from that we can conclude that the line $NP$ passes through $O$.

  • 1
    $\begingroup$ I appreciate very much your answer. $\endgroup$
    – Jean Marie
    Nov 12, 2021 at 18:16

Any four distinct points A, B, C, D can be inverted into the vertices of a parallelogram $A'B'C'D'$ (including, as one possibility, a degenerate parallelogram in which the four vertices lie on one line, but still $A'B' = D'C'$ and $A'D' = B'C'$). Hint: Consider separately the three cases (i) $AC // BD$; (ii) $AB // CD$ or $AD // BC$; (iii) A, B, C, D not concyclic.

Note: the // notation refers to point-pair separation. One way of characterizing is that AC // BD when every circle through A and C intersects every circle through B and D.

Finally I actually asked a question Is this usage of the term 'Jacobian' related to the other uses e.g. matrix or elliptic function? related to this (where this is repeated) that I never had anyone answer!

The diagrams below give my rendition of the book (Geometry Revisited) solution for the (iii) case.

Exercise 5.7.6

5.7.6 supplementary

5.7.6 Lemma

  • $\begingroup$ Interesting property I wasn't aware of ! $\endgroup$
    – Jean Marie
    Feb 28, 2019 at 10:59

The general Soddy hexlet:

can be obtained by an inversion of the particular Soddy hexlet with 6 equal spheres (called the annular hexlet on Wikipedia):

In fact this also works for an arbitrary number of spheres, not only six. The envelop of the spheres is a Dupin cyclide. Example for 5 spheres:

The Villarceau circles of the cyclide are displayed on the above image. Here is an interesting property of the objects shown on this image.

  • $\begingroup$ Very captivating. Thank you. One of the references given in your link is to a paper where one of the authors, Lionel Garnier, is somebody I know, specialized on Dupin's cyclides and their connections with Bezier curves in the framework of computer graphics. $\endgroup$
    – Jean Marie
    Nov 30, 2022 at 14:59

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