Linear transformations in two dimensions are defined by transforming lines into lines. If we look at things more in a polar-coordinates, complex-numbers way, we find another set of transformations which transform shapes of a particular simple sort into shapes of the same sort.

Circles are in some ways for polar coordinates what lines are for cartesian coordinates. The new set of transformations is represented in complex numbers by equations like:

w=(az+b)/(cz+d)

All transformations like that transform circles into circles – so long as we interpret lines as special “circles” with centre at infinity and infinite radius. In the same sort of way, some linear transformations in two dimensions transform lines into special “lines”, i.e. points.

**If the locus of z is a circle (or line), then the locus of w=(az+b)/(cz+d) is a circle (or line). A diameter of the locus of z passing through the point z=d/c transforms into a diameter of the locus of w.**

**Proof:** If the locus of w+k is a circle (or line), then the locus of w is a circle (or line), just shifted by k. If the locus of mw is a circle (or line), then the locus of w is a circle (or line), just shifted and, if it’s a circle, made smaller or larger.

So it’s enough to consider the transformation w=1/z. If that transforms circles (or lines) into circles (or lines), then all transformations of form w=(az+b)/(cz+d) do that too.

We go by way of looking at the transformation 1/z*. If z = r cis θ, then 1/z* = (1/r) cis θ. 1/z* is in the same direction as z, but (modulus of z)×(modulus of (1/z*))=1

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First take the case where the origin is inside the circle. (This is called the “Intersecting Chords Theorem”).

Let origin = P in the diagram above. Geometry tells us that in the diagram above, AP·PD = BP·PC, in other words AP.PD is constant whatever point A we choose on the circumference if D is the other end of the chord through P. Thus for some constant k, as z moves round the circle, the reflection in the imaginary axis of k/z* moves round the same circle; so k/z* moves round a circle; so 1/z* moves round a circle; and so 1/z does too. And a diameter going through P transforms into a diameter.

The geometry: triangles ABP and CDP are *similar* because:

∠BAD = ∠BCD, as inscribed angles subtended by the same chord BD,

∠ABC = ∠ADC, as inscribed angles subtended by the same chord AC,

∠APB = ∠CPD, as a pair of vertical angles.

So AP/BP = DP/CP, and thus AP·PD = BP·PC ▇

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If the origin D (which is the pole, or centre of inversion) is outside the circle, as above, then there is a similar result, the “Intersecting Secants Theorem”.

CLAIM: if z moves on a circle, the tangent from the origin to the circle has length k, and m=^{2}, then w=m/z* moves on the same circle, and the diameter of the z-circle ⟼ diameter of the w-circle.

PROOF: A is any point on the circle; E is the centre; F is the midpoint of CA, so angle CFE is a right angle. D is the origin.

Because DFE and CFE are both right-angled triangles

Pythagoras ⇒ DE^{2}−DF^{2}=EF^{2}=CE^{2}−CF^{2}

⇒ DE^{2}−CE^{2}=DF^{2}−CF^{2} [*]

But DC.DA = (DF−CF).(DF+FA) = (DF−CF).(DF+CF) = DF^{2}−CF^{2}

and DG.DH = (DE−GE).(DE+EH) = (DE−CE).(DE+CE) = DE^{2}−CE^{2}

since GE, CE, and EH are all radii of the circle

So equation [*] ⇒ DC.DA = DG.DH. Since this is true wherever A is on the circle, it is also true when A is at B, and so DC.DA = DG.DH = DB2 = k^{2}

So DC is in the same direction as DA, but of length (or magnitude, or modulus) equal to k^{2}/|DA|. If A is represented by the complex number z, C is represented by the complex number k^{2}/z*

Since the whole transformation is symmetrical around the line DH, the diameter GH of the z-circle ⟼ the diameter HG of the w-circle (though in general other diameters of the z-circle do not ⟼ diameters of the w-circle). ▇

The final case is when the origin (pole) is on the circle, i.e. neither inside the circle nor outside it. When the pole is *very close* to the original circle, then the image circle is very big. So we’d expect that when the origin (pole) is actually on the original circle, the image will be a *line*. So it is. We have one of the diagrams below, where D is the origin and A is such that DA is the diameter, with DA.DA'=DB.DB'

∠ADB is common to the triangles ΔABD and B'A'D, and DA/DB=DB'/DA'. Therefore the triangles are similar.

Therefore ∠B'A'D = ∠ABD = π/2, since ∠ABD is an angle in a semicircle.

Therefore the locus of B', i.e. of 1/z*, if B represents z, is a straight line perpendicular to the diameter of the circle through D and distant 1/d from D where d=diameter.

Conversely, if B' represents z, moving on a line with perpendicular distance a from D, this reasoning shows us that the locus of 1/z* is a circle going through D, with diameter perpendicular to the locus of z, and diameter 1/a. The two ends of the diameter are the images under the transformation of infinity and of the point on the line closest to D.

The final case is when z moves along a line passing through the origin D. 1/z* has the same argument as z, and its modulus takes all real values (including 0; z→∞ ⟼ 1/z*→0), so the locus of 1/z* is the same line.

In short:

If z moves on a circle not passing through the origin, m/z* moves on the same circle, for some positive constant m.

If z moves on a circle passing through the origin of diameter d, 1/z* moves on a line perpendicular to the diameter through the origin.

If z moves on a line not passing through the origin, 1/z* moves on a circle passing through the origin.

If z moves on a line passing through the origin, 1/z* moves on the same line.

A transformation like w=(az+b)/(z+d) (I’ve divided through by a common factor to make c=1) can be broken down into successive transformations:

z ⟼z+d (translation)

z+d ⟼ 1/(z+d)* (inversion+conjugate)

1/(z+d)* ⟼ 1/(z+d) (inversion)

1/(z+d) ⟼ (b−ad) [1/(z+d)] (enlargement and rotation)

(b−ad) [1/(z+d)] ⟼ a + (b−ad) [1/(z+d)] = (az+ad+b−ad) / (z+d) = (az+b) / (z+d) (translation)

We can calculate the image by drawing each transformation in turn.

**But there’s an easier rule of thumb:**

You identify whether the locus of w is a circle or a line by seeing whether z takes a value which will make w go to infinity (so its locus is a line), or doesn’t.

Then you identify the w-circle by finding a diameter, given by the w-values got from z-values at two ends of a diameter along the line connecting the centre of the z-circle from the centre of inversion (z=−d/c).

Or, if z moves on a line, a diameter of the w-circle is given by the w-values got from z at infinity and from z at the closest point on the line to the centre of inversion.

You identify the w-line by using any two convenient z-values to calculate two w-values.

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**Alternatively:** if the locus of z is |z|=k, to identify either two ends of a diameter of the w-circle, or the two points from which the w-line is equidistant, do the following.

Rearrange the equation w=(az+b)/(cz+d) and we get z=(−b+dw)/(a−cw), or: the distance of w from b/d is k|c|/|b| times the distance of w from a/c.

Let λ=k|c|/|b|. Look at the line connecting b/d and a/c. Unless λ=1, there will be two points on that line which have (distance from b/d)=λ(distance of w from a/c), one (call it M) between b/d and a/c and one (call it L) outside.

The circle must be symmetrical around LM, because if a point on one side fits the equation

(distance from b/d)=λ(distance of w from a/c)

then the reflection of that point in LM also fits the equation.

Therefore LM is the diameter of the circle.

See here for another, more traditional, proof of this result, which is called the “Circle of Apollonius”.