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With determinant-one normalization, the necessary and sufficient condition for sphere rotations as special-unitary Möbius transformations isEquivalently, the matrix $M=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ belongs to the special unitary group $SU(2)$. The two determinant-one representatives $M$ and $-M$ induce the same Möbius transformation, and both satisfy this condition.

Use north-pole stereographic projection in the direction from the plane to the sphere:Its chord-distance identity is $|\phi (z)-\phi (w)|=2\chi (z,w)$, where $\chi$ is the chordal metric. Direct subtraction, using $ad-bc=1$, givesThis is first calculated away from poles and then extended continuously to the Riemann sphere. A sphere rotation preserves chord distances, so for distinct finite nonpole pointsUsing three distinct points shows all these positive ratios equal one. Hence $H(z)=1+|z{|}^2$ everywhere. Comparing its coefficients givesThus $M$ is a unitary matrix. Comparing $M^{-1}=M^{\ast }$ with $M^{-1}=\left(\begin{array}{cc}d& -b\\ -c& a\end{array}\right)$ proves the boxed relations.

Conversely, these relations give $H(z)=1+|z{|}^2$, so the induced sphere map $T=\phi g{\phi }^{-1}$ preserves all Euclidean chord distances. Such a bijection of the unit sphere is the restriction of an orthogonal matrix: let $v_j=T(e_j)$; preservation of distances makes these an orthonormal basis, and $T(p)\cdot v_j=p\cdot e_j$ forces $T(p)={\sum }_j{p}_j{v}_j$. Finally the Möbius transformation is holomorphic with positive real Jacobian at every ordinary coordinate point. Conjugating by stereographic projection preserves orientation, so this orthogonal matrix has determinant $+1$. It is therefore a genuine rotation in three dimensions, proving sufficiency rather than just spherical isometry.

For the circle assertion, a nondegenerate generalized circle has an equationFor $A\ne 0$ this is an ordinary circle, and for $A=0$ it is a line with infinity included. On the sphere write coordinates $(p_1,p_2,p_3)$, with $z=(p_1+i{p}_2)/(1-p_3)$ and $|z{|}^2=(1+p_3)/(1-p_3)$. Multiplication by $1-p_3$ gives the planeIf $n=(2\mathrm{Re}B,-2\mathrm{Im}B,A-C)$ and $h=-(A+C)$, then $|n{|}^2-h^2=4(|B{|}^2-AC)>0$. Thus the plane intersects the unit sphere in a proper circle. It contains the north pole exactly when $A=0$, accounting for the point at infinity on a line.

Conversely, a circle on the sphere is the intersection with a plane $n\cdot p=h$, $|h|<|n|$. Set $A=(n_3-h)/2$, $C=(-n_3-h)/2$ and $B=(n_1-i{n}_2)/2$. The preceding equation is recovered and has $|B{|}^2-AC=(|n{|}^2-h^2)/4>0$. Its inverse stereographic image is an ordinary circle or a line with infinity attached. This proves both directions of the circle-plane relation under stereographic projection.

Use the unit sphere $X^2+Y^2+Z^2=1$, its north pole $(0,0,1)$, and the equatorial complex plane $Z=0$. The line from the north pole to a sphere point meets that plane atThe north pole maps to $\infty$. The inverse stereographic projection isThe antipodal point has coordinate $w=-(X+iY)/(1+Z)$. Since $X^2+Y^2=(1-Z)(1+Z)$, this is $-1/\bar{z}$ when both sides are finite and nonzero. Zero and infinity correspond to the two poles and are exchanged by the same extended convention. The inverse formula also proves the converse, giving the antipodal stereographic coordinate relation

The requested rotation result is that sphere rotations correspond exactly to Möbius transformationsThe matrices $\left(\begin{array}{cc}a& b\\ -\bar{b}& \bar{a}\end{array}\right)$ form $SU(2)$, and opposite matrices induce the same transformation. Thus the rotation group $SO(3)$ is identified with $SU(2)/\{\pm I\}$. This is the sphere rotations as special-unitary Möbius transformations result, which need not be proved here.

For a planar circle $|z-c|=r$, $c=a+ib$ and $r>0$, substitute $|z{|}^2=(1+Z)/(1-Z)$ and $\Re (\bar{c}z)=(aX+bY)/(1-Z)$. Multiplication by $1-Z$ gives the circle-plane relation under stereographic projection:The lift lies in this plane, and conversely its intersection with the sphere projects onto the circle. It is a nondegenerate circle on the sphere. The north pole is not on the displayed plane, since substitution there gives two rather than zero.

A great circle is the intersection with a plane through the sphere's centre. Here this holds exactly whenThe unit circle lifts to the equator. If a different circle lifts to a great circle, its plane and the equatorial plane are distinct planes through the origin, so their intersection cuts the sphere at two antipodal equatorial points. Their planar coordinates are $z$ and $-z$ on $|z|=1$.

Conversely, if the planar circle meets the unit circle in $z$ and $-z$, their lifts are antipodal equatorial points. The plane of the lifted circle contains their midpoint, the origin, so that circle is great. Equivalently, adding $|z-c{|}^2=r^2$ and $|-z-c{|}^2=r^2$ gives $r^2=1+|c{|}^2$. The exclusion of the unit circle ensures the two intersection points are a genuine pair rather than coincident circles. This proves the antipodal equatorial intersection criterion for a great circle.