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Fix $p$ on a Riemannian surface. For sufficiently small $r>0$, the geodesic circles of radius $r$ centered at $p$ and at every sufficiently nearby distinct $q$ meet transversally at exactly two points. In a strongly convex normal neighbourhood, write $q={\exp }_p(\epsilon v)$ and $x(\theta )={\exp }_p(re(\theta ))$. First variation givesThe quotient of the bracketed function by $\epsilon$ extends smoothly and has two simple zeros at zero. The implicit function theorem, uniformly over unit $v$, preserves these two zeros and excludes others. The nonzero angular derivatives prove transversality. The distinct-centre hypothesis is essential: coincident circles are not transverse.

For $r$ smaller than the normal-coordinate radius, the geodesic circle on a Riemannian surface consists of points at distance $r$ from $p$. The exponential map restricted to the radius-$r$ circle in $T_{p}S$ gives its smooth embedded parameterization. Outside the injectivity range a distance sphere need not be smooth.

The literal assertion needs $q\ne p$. If $q=p$, the two geodesic circles coincide, have equal tangent lines at every point, and do not intersect transversally; their intersection is not finite. We prove that close geodesic circles intersect twice for nearby distinct centres, and in fact find exactly two intersections.

Choose a strongly convex normal neighbourhood about $p$ and a sufficiently small $r>0$ so that every circle considered and all geodesics joining its points to nearby centres remain there. Squared distance $d(q,x{)}^2$ is then smooth in both arguments, and the exponential map is nonsingular on the relevant radius-$r$ tangent circles. Consequently all these distance spheres are smooth embedded one-dimensional manifolds.

Write $q={\exp }_p(\epsilon v)$ with $|v|=1$ and $\epsilon >0$, and parameterize $S_r(p)$ by $x(\theta )={\exp }_p(re(\theta ))$. DefineThe first variation of geodesic energy givesIndeed differentiating the energy of the joining geodesic cancels the interior term by the geodesic equation, leaving the initial endpoint term $-2⟨re(\theta ),v⟩$. This is also the endpoint form of the Gauss lemma.

Since $F(0,v,\theta )=0$, the quotient $F/\epsilon$ extends smoothly to $\epsilon =0$ by integrating ${\partial }_{\epsilon }F$ from zero to $\epsilon$. At zero it is $-2r\cos (\theta -\arg v)$, with exactly two simple zeros. The implicit function theorem gives two nearby simple zeros for small positive $\epsilon$. Compactness of the unit vector circle makes the choice of small $\epsilon$ uniform in $v$; away from fixed neighbourhoods of those two zeros the cosine is bounded away from zero, so there are no additional zeros.

A nonzero angular derivative at either zero says that the defining function of $S_r(q)$ has a nonzero derivative along $S_r(p)$. Thus their tangent lines differ: the intersections are transverse. Shrinking the neighbourhood of centres accordingly proves

For $v\in T_{p}S$ for which the geodesic exists up to time one, the exponential map is ${\exp }_p(v)={\gamma }_v(1)$, where ${\gamma }_v(0)=p$ and ${\dot{\gamma }}_v(0)=v$. The initial velocity determines the geodesic uniquely. Since $d({\exp }_p{)}_0$ is the identity, this map is a local diffeomorphism near zero.

Choose an orthonormal basis of $T_{p}S$ and set $e(\theta )=(\cos \theta ,\sin \theta )$. The geodesic polar coordinates arein a normal neighbourhood, away from its centre. For sufficiently small positive $r$, the geodesic circle iswithin that neighbourhood. Outside the injectivity range the distance sphere need not have this smooth coordinate description.