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Normalize three distinct points on a generalized circle by a Möbius transformation $f$ taking them to $1,0,\infty$. The inverse image of the extended real line is their generalized circle, and $J=f^{-1}\circ c\circ f$, with $c(z)=\overline{z}$, fixes that circle pointwise and squares to the identity. Changing the three normalizing points replaces $f$ by $M\circ f$ with a real-coefficient Möbius map $M$. Since $M$ commutes with conjugation, $J$ is unchanged. Thus this construction depends only on the circle. It is inversion in a circle for an ordinary circle and Euclidean reflection for a straight line, each extended to the Riemann sphere.

Inversion in a circle of centre $a$ and radius $R$ is an orientation-reversing involution of the Riemann sphere fixing that circle pointwise and exchanging its interior and exterior. It is an anti-Möbius transformation. Reflection in a line is the corresponding inversion for a generalized circle through infinity. A composition of two such inversions is a Möbius transformation.

For a Euclidean circle with centre $c$ and radius $r$, two distinct finite points are inverse when they lie on the same ray from $c$ and their distances from $c$ have product $r^2$. Points on the circle are fixed, and $c$ and infinity are inverse to each other. In complex coordinates this uniquely defines inversion in a circle byUniqueness follows from the ray and product conditions, including the specified centre–infinity pair; the formula also gives $I_{\Gamma }^2=\mathrm{id}$. A circle through infinity is a straight line in the complex plane; its inversion is reflection in that line, fixing infinity. For a line through $c$ with direction angle $\theta$, its formula is $c+e^{2i\theta }\overline{z-c}$.

Thus every inversion is an anti-Möbius transformation, of the form $(a\overline{z}+b)/(c\overline{z}+d)$ with nonzero determinant. If $J(z)=\overline{z}$, write such an inversion as $M\circ J$ with $M$ a Möbius transformation. Since $J\circ M\circ J$ is the Möbius transformation with conjugated coefficients, the product of two inversions is $M_1\circ (J\circ M_2\circ J)$, hence Möbius. Pairing consecutive inversions proves every composition of an even number of inversions is a Möbius transformation.

inversion in a circle with center $c$ and radius $r>0$ is $I(z)=c+r^2/(\overline{z}-\overline{c})$, with $c$ and infinity interchanged. It fixes the circle pointwise and reverses complex orientation. Each inversion is fractional-linear in $\overline{z}$, so composing two eliminates the conjugation and gives $(az+b)/(cz+d)$ with nonzero determinant: a Möbius transformation.

Translate and rotate two distinct centers to $0,d$, where $d>0$, and let the radii be $r,s$. The composition of inversion in the first circle followed by the second isIts fixed points solve $d{z}^2-(d^2+r^2-s^2)z+d{r}^2=0$. The discriminant isDisjoint circle boundaries require either $d>r+s$ or $d<|r-s|$, so both factors have the same nonzero sign and $\Delta >0$. There are two distinct fixed points on the line of centers. If the circles are concentric, their radii differ and the composition is $z\mapsto (s^2/r^2)z$, with exactly the two fixed points $0,\infty$. Thus the composition has two distinct fixed points, also for nested disjoint circles. Reversing the order inverts the Möbius transformation and leaves its fixed points unchanged.

Use the existing cross-ratio conventionValues involving infinity are defined by limits. More uniformly, represent each point by a nonzero homogeneous vector $u=(u_0,u_1)$ and put $D(u,v)=u_0{v}_1-u_1{v}_0$. Then the same value is $D(z,a_1)D(a_0,a_2)/[D(z,a_2)D(a_0,a_1)]$, interpreted in the Riemann sphere. Rescaling any representative cancels out. The first three points are distinct, so numerator and denominator cannot simultaneously vanish. The fourth point may coincide with one of them, giving the values $1,0,\infty$ at $a_0,a_1,a_2$, respectively.

With $a_0,a_1,a_2$ fixed, this is a nonconstant Möbius transformation $f$. For finite first three points, write $C=(a_0-a_2)/(a_0-a_1)\ne 0$; thenThe determinant of this fractional-linear map is nonzero because $a_1\ne a_2$. Its inverse is defined at every point of the sphere, including poles and infinity, so each $w$ has exactly one preimage. The same conclusion follows directly from the homogeneous formula when an $a_i$ is infinity.

Möbius transformations preserve generalized circles. To verify the fact used here, affine complex changes preserve ordinary circles and lines, while a generalized circle has an equation $A|z{|}^2+Bz+\overline{B}\overline{z}+D=0$ with $A,D$ real. Substitution $z=1/w$ and multiplication by $|w{|}^2$ gives another equation of the same form. Every non-affine Möbius map is a composition of affine maps and $z\mapsto 1/z$, so the claim follows. Consequentlyis a generalized circle. It contains the three distinct $a_i$, so it is their unique ordinary circle, or their line together with infinity when they are collinear. This is the real cross-ratio criterion for a generalized circle.

Let $c(w)=\overline{w}$, with $c(\infty )=\infty$. The defining equation for $J$ now gives the cross-ratio construction of generalized-circle reflection:This is well-defined by bijectivity of $f$, and $J^2=f^{-1}{c}^{2}f$ is the identity. Moreover $J(z)=z$ precisely for $z\in S$.

If the first three points lie on the extended real line, the normalization map has real coefficients, so $f(\overline{z})=\overline{f(z)}$. Hence $J(z)=\overline{z}$. For a second triple $b_0,b_1,b_2$ on the same generalized circle, write its normalization as $g=Mf$. Each $f(b_i)$ is real or infinity, and the unique Möbius map taking this real triple to $1,0,\infty$ has real coefficients. Thus $M$ commutes with $c$, givingTherefore $J$ depends only on the generalized circle, not on the three selected points. For an ordinary circle with centre $q$ and radius $R$, it is inversion in a circle, $J(z)=q+R^2/(\overline{z}-\overline{q})$; for a straight line it is the corresponding Euclidean reflection, extended to infinity.

An inversion in a circle on the Riemann sphere is the anti-Möbius involution fixing that generalized circle pointwise. For a Euclidean circle with centre $a$ and radius $R>0$, it isIt reverses the argument relative to the centre and sends radius $r$ to $R^2/r$, so exactly the points of the circle are fixed. For a straight line, regarded as a circle through infinity, inversion is the ordinary reflection in that line. These formulas prove existence. For uniqueness, the composition of two anti-Möbius transformations fixing the same circle pointwise is a Möbius transformation fixing at least three distinct points, hence the identity. Therefore the two inversions agree.

Every inversion has the form $(a\overline{z}+b)/(c\overline{z}+d)$ with $ad-bc\ne 0$. Composing two conjugate-linear fractional expressions cancels the complex conjugations and gives $(Az+B)/(Cz+D)$ with nonzero determinant. Induction proves that an even number of inversions gives a Möbius transformation.

Conversely translations are compositions of reflections in parallel lines, rotations are compositions of reflections in two lines through their centre, and positive dilations are compositions of two concentric circle inversions: $J_R{J}_1(z)=R^{2}z$. Thus every nonconstant affine map $z\mapsto az+b$ is an even composition. The map $z\mapsto 1/z$ is $J_{|z|=1}$ composed with reflection in the real axis, also an even composition. If $c\ne 0$, factor a general Möbius transformation asThis is a composition of affine maps and $1/z$. If $c=0$, it is already affine. Every Möbius transformation is therefore a composition of an even number of inversions, with the identity represented by zero inversions or by any inversion repeated twice.

For speed $v=k{r}^2$ in the punctured plane, the travel-time element is $ds/(k{r}^2)$. Under the inversion in a circle with radial coordinate $u=1/r$, this becomes $k^{-1}\sqrt{d{u}^2+u^{2}d{\theta }^2}$, the arc length in Euclidean space divided by $k$. Thus a shortest travel-time route corresponds to a straight line segment in the inverted plane, whenever that segment avoids the origin and lies in the allowed region. Between $(a,0)$ and $(0,a)$ the inverted segment has equation $u(\cos \theta +\sin \theta )=1/a$. The route is $r=a(\cos \theta +\sin \theta )$, and its minimum travel time is $\sqrt{2}/(ka)$. The triangle inequality for arc length proves global minimality.