= Solution
With determinant-one normalization, the necessary and sufficient condition for <sphere rotations as special-unitary Möbius transformations> is
$$
\boxed{c=-\overline b,\qquad d=\overline a,\qquad |a|^2+|b|^2=1.}
$$
Equivalently, the matrix $M=\begin{pmatrix}a&b\\c&d\end{pmatrix}$ 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>:
$$
\varphi(z)=\frac{(2\operatorname{Re}z,2\operatorname{Im}z,|z|^2-1)}{1+|z|^2},\qquad
\varphi(\infty)=(0,0,1).
$$
Its chord-distance identity is $|\varphi(z)-\varphi(w)|=2\chi(z,w)$, where $\chi$ is the <chordal metric>. Direct subtraction, using $ad-bc=1$, gives
$$
\chi(g(z),g(w))=
\frac{|z-w|}{\sqrt{H(z)H(w)}},\qquad
H(z)=|az+b|^2+|cz+d|^2.
$$
This 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 points
$$
\frac{H(z)}{1+|z|^2}\frac{H(w)}{1+|w|^2}=1.
$$
Using three distinct points shows all these positive ratios equal one. Hence $H(z)=1+|z|^2$ everywhere. Comparing its coefficients gives
$$
|a|^2+|c|^2=1,\quad |b|^2+|d|^2=1,\quad
a\overline b+c\overline d=0.
$$
Thus $M$ is a <unitary matrix>. Comparing $M^{-1}=M^*$ with $M^{-1}=\begin{pmatrix}d&-b\\-c&a\end{pmatrix}$ proves the boxed relations.
Conversely, these relations give $H(z)=1+|z|^2$, so the induced <sphere> map $T=\varphi g\varphi^{-1}$ 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 $T(p)=\sum_jp_jv_j$. 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 equation
$$
A|z|^2+Bz+\overline B\,\overline z+C=0,
\qquad A,C\in\mathbb R,\quad |B|^2-AC>0.
$$
For $A\ne0$ 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 $z=(p_1+ip_2)/(1-p_3)$ and $|z|^2=(1+p_3)/(1-p_3)$. Multiplication by $1-p_3$ gives the plane
$$
2\operatorname{Re}B\,p_1-2\operatorname{Im}B\,p_2+(A-C)p_3+(A+C)=0.
$$
If $n=(2\operatorname{Re}B,-2\operatorname{Im}B,A-C)$ 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 $B=(n_1-in_2)/2$. 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>.
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