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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.