Then M = M' iff OM = R : the set of all fixed points in inversion is the

Consider intersection points A and B of OM with the inversion circle. OM.OM' = OA² = OB² says that (ABMM') is an harmonic range.

An inversion by a positive power R² is the same as :
M' is the harmonic conjugate of M, on line OM, with respect to the inversion circle (Or in other word, M' is the foot of polar of M) |

A negative power inversion may be considered as an inversion with power +R², and a central symmetry from O (dilation with ratio -1).

Some then consider only positive power inversion, discarding a generalisation to any power.

We consider here the general case : any power, positive or negative.

Let ℑ

OM"/OM = k

Note that ℑ

Inversion is

Points M and M' are then

Product of an inversion with power k by a dilation with ratio r is an inversion with power r.k

We can't say anything about composition of inversions with different poles.

By going to limit from a secant, we get : if two curves are tangent in M, their inverses are tangent in M', inverse of M.

This is really a specific case of something more general :

If two curves intersect in M with angle α (angle of tangents)
their inverses intersect in M', inverse of M, with angle π - α |

Really, near points M and M', a curve and its inverse (their tangents) are symmetric
through the perpendicular bisector of MM' :

Let MP a secant line to curve Γ and M'P' the corresponding secant to Γ'
inverse of Γ, that is M and M' are inverses of each other, similarily for P and P'.

Then k = OM.OM' = OP.OP'
says that MM'PP' are on a same circle. Let (C) the circle MPM'P'.
Going to the limit, secants become the tangents in M to Γ and in M' to Γ'.

Circle (C) → (γ) with the same tangents in M and M', hence these tangents
are symmetric through the perpendicular bisector of chord MM' in that circle (γ).

A further consequence is if two curves are perpendicular in M, their inverses are perpendicular in M', inverse of M (π - π/2 = π/2).

Exactly : points A,B,A',B' being concyclic, triangles OAB and OB'A' are similar.

That is : A'B'/BA = OA'/OB = (OA.OA') / (OA.OB) = |k|/(OA.OB)

A'B' = AB × |k|/(OA.OB)

If O,A,B are in line, A' and B' are on this same line and we directly get the relation.

Even more precise : A'B'/AB = -k/(OA.OB)

If the line doesn't go through O, let A the orthogonal projection of O on the line, and A' its inverse.

Let M any point on the line and M' its inverse.

OM.OM' = OA.OA' says that MM'AA' are concyclic, hence angle MM'A' is a right angle.

Triangle OM'A' being right, locus of M' is the circle with diameter OA'

The inverse of a line (d) not going through pole O is
a circle centered in ω and going through the pole (and reciprocally). Oω is perpendicular to (d) |

From relation OA.OA' = k and OA' = 2Oω, we get
k = (2OA).Oω = OO'.Oω

The center ω of the circle is the inverse of the reflection O' of the pole through the line.

Note : When the line and the circle intersect, the intersection points are fixed points,
hence lie on the inversion circle (and power is positive).

Let A and B the
ends of diameter in line with O, A' and B' their inverses.
M any point on the circle, M' its inverse. N the other intersection point of OM with the circle and N' its inverse.

Then AA'MM' are concyclic, or in other words AM and A'M'
are antiparallel with respect to OM and OA.

Also in circle (C), AM and NB are antiparallel with respect to the same lines.
Hence M'A' is parallel to NB.
M' is then the transformed of N in a dilation centered in O with ratio OA'/OB.

When M, hence N, walks the circle (C), M' walks the homothetic circle (C')

The inverse of a circle not going through pole is an homothetic circle. |

A'B' = 2R' being the inverse of AB = 2R, we get : R' = R× |k|/(OA.OB) and because OA.OB is the power of O relative to (C), noted C(O)

R' = R |k/C(O)|

As inversion is involutive, (C) is inverse of (C') and R = R' |k/C'(O)|, hence :

R'²/C'(O) = R² /C(O) et k² = C(O).C(O')

Note that the center ω' of the inverted circle is **not** the inverse of ω

Really, Oω' = (OA'+OB')/2 = k/2 (1/OA + 1/OB)

Let P the harmonic conjugate of O with respect to A,B (P is the foot of the polar of O to circle (C)) :
2/OP = 1/OA + 1/OB, hence :
Oω' = k/OP

The center of inverse circle is the inverse of the foot of polar of O

A circle is globally unchanged iff k = C(O)

When k positive, this is equivallent to :

OM.OM' = OT² = C(O), OT perpendicular to Tω.

A circle is globally unchanged iff it is perpendicular to the inversion circle (k>0)

The two circles inverse of each other are also homothetic from the inversion pole, hence there are two inversions which exchange two given circles, the poles are the two homothecy centers I and J of the two circles. If the circles are tangent, the touch point is one homothecy center, but it must be rejected because it lies on the circles, and corresponding inversion transforms the circles into two parallel lines. In that case, there is only one inversion exchanging the two circles.

We may then suitably choose the inversion so as to simplify the problem, changing a circle into a straight line, any two circles into concentric circles etc...

Of course not intersecting circles ! (inversion keeps intersection properties)

Let Δ the radical axis of the two circles (locus of points with same power to both circles). P any point on the radical axis. P is the center odf a circle Γ perpendicular to both given circles : PT

Γ intersects the center line in I and J which are the limit points of the circles pencil defined by C

An inversion from pole I transforms Γ into a line d, and circles C

Let ω the inverse of J (that is intersection of d with the center line).

The pencil of lines d, going trough ω is then made of lines, each perpendicular to C'

Here the inversion power has been choosen such as to keep circle C'

C'

Note : Here C_{2} is inside C_{1}, but the construction also works
if C_{2} is some circle of the pencil "on the other side" of radical axis.

This results into constructing an "hyperbolic" Steiner chain :

The two movable blue points O and r allow to zoom and pan.

The "trivial" chain in concentric circles (cyan) is between circles with radii r and R = 3r/2.

Moving point P manually rotates the chain, when not running.

Moving point I is the inversion pole.
Inversion keeps the outside circle of the "trivial" chain unchanged, inverting everything else.

If the pole is inside the small circle, or outside the large circle, it results into a "normal" Steiner chain.

Set as at start of applet, the inverse of the small circle is in the "other side" part of the pencil,
resulting into an "hyperbolic" chain (centers of red circles lie on an hyperbola).

In green, construction of inverses of touch points and cyan circles → red circles :

Let T the touching point (trivial, on OP) of circle centered in P with the large circle.

Then T' is the other intersection point of IT with the large circle (globally unchanged).

The center of the inverse red circle is then the intersection point of OT' and PI.

This construction fails when these two lines are parallel
(red circle becoming a straight line, as the cyan circle goes through I), or overlap
(center of circle is on OI).

This can be seen by manually moving P carefully.

In both cases, we must use some other construction,
which can't be done automatically by the applet : the red circle vanishes, or becomes wrong.