content_image/310672.jpg content_image/310673.jpg
γ at s $^{+}$( s ) and s $^{−}$( s ) . 

## Proof of Claim 2: Computing, we get 

<img src='content_image/310672.jpg'>

Since ν ′ 2 ≡ ∂ν$_{2}$ ∂r and ν ′ 1 ≡ ∂ν$_{1}$ ∂r are the only unknowns, we end up having to invert 

<img src='content_image/310673.jpg'>

again and this is always nonsingular, giving us ν ′ 1 and ν ′ 2 as a function of s, the coordinate of the center of the disk. 

Relative to the horizontal, the angle of the curve at s $^{+}$( s ) is θ$_{1}$ + ν$_{1}$ so the rate of change in angle as we expand the circle is ∂ θ$_{1}$ ∂r + ν ′ $_{1}$. Recalling that rate of movement of this exit point as we expand the circle is given by ∂ k ∂r = sec ν$_{1}$ , we have that the curvature is given by κ ( s $^{+}$( s )) = ∂ k $_{∂r}$( ∂ θ$_{1}$ ∂r + ν ′ $_{1}$) = sec ν$_{1}$ ( ∂ θ$_{1}$ ∂r + ν ′ $_{1}$). Similarly, κ ( s $^{−}$( s )) = sec( ν$_{2}$ )( ∂ θ$_{2}$ ∂r + ν ′ $_{2}$). □ 

Claim 3. Generically, we can deduce s $^{+}$( s ) from knowledge of ν$_{1}$ ( s ) , ν$_{2}$ ( s ) , θ$_{1}$ ( s ) and θ$_{2}$ ( s ) . 

Proof: We outline the proof without some of the explicit constructions that follow without much trouble from the outline. We have that θ$_{1}$ ( s $^{−}$( s )) + ν$_{1}$ ( s $^{−}$( s )) = π − θ$_{2}$ ( s ) − ν$_{2}$ ( s ) and θ$_{1}$ ( s ) + ν$_{1}$ ( s ) = π − θ$_{2}$ ( s $^{+}$( s )) − ν$_{2}$ ( s $^{+}$( s )). All four of these quantities (the left- and right-hand sides of each of the 2 equations) are the turning angles between the tangent to the curve at the center of the disk and the tangent to the curve at a point 

