Well, we have
$(\alpha - c) \cdot (\alpha - c) = r^2, \tag 1$
since $\alpha$ lies on the sphere of radius $r$ centered at $c$; this is just what equation (1) affirms. If we differentiate (1) with respect to $s$, the arc-length along $\alpha$, we obtain
$2\alpha' \cdot (\alpha - c) = \alpha' \cdot (\alpha - c) + (\alpha - c) \cdot \alpha' = 0, \tag 2$
whence
$\alpha' \cdot (\alpha - c) = 0; \tag 3$
since $\alpha$ is a unit-speed curve, we have
$\alpha' = T, \; \Vert T \Vert = 1, \tag 4$
the unit tangent vector to $\alpha$; thus (3) becomes
$T \cdot (\alpha - c) = 0, \tag 5$
which is nearly self-evident, since $T$ is tangent to the sphere (1), hence normal to the radial vector $\alpha - c$; in any event, we may differentiate (5) once again to obtain
$T' \cdot (\alpha - c) + T \cdot T = T' \cdot (\alpha - c) + T \cdot \alpha' = 0, \tag 6$
or
$T' \cdot (\alpha -c) + \Vert T \Vert^2 = 0; \tag 7$
we now use the Frenet-Serret equation
$T' = \kappa N, \tag 8$
where $N$ is the normal vector to $\alpha$, and (4) to re-write (7) as
$\kappa N \cdot (\alpha - c) + 1 = 0, \tag 9$
which since $\kappa > 0$ implies
$N \cdot (\alpha - c) = -\dfrac{1}{\kappa}; \tag {10}$
we may now differentiate (9) with respect to $s$ to find
$\kappa' N \cdot (\alpha - c) + \kappa N' \cdot (\alpha - c) + kN \cdot \alpha'= 0; \tag{11}$
we have
$N \cdot \alpha' = N \cdot T = 0, \tag{12}$
and also the Frenet-Serret equation
$N' = -\kappa T + \tau B, \tag{13}$
where $\tau$ is the torsion and $B = T \times N$ the binormal vector of $\alpha$; when (10), (12) and (13) are substituted into (11) we obtain
$-\dfrac{\kappa'}{\kappa} + \kappa (-\kappa T + \tau B) \cdot (\alpha - c) = 0, \tag{14}$
or
$-\dfrac{\kappa'}{\kappa} -\kappa^2 T \cdot (\alpha - c) + \kappa \tau B \cdot (\alpha - c) = 0, \tag{15}$
which by virtue of (5) reduces to
$-\dfrac{\kappa'}{\kappa} + \kappa \tau B \cdot (\alpha - c) = 0, \tag{16}$
and since $\kappa > 0 \ne \tau$ we have
$-\dfrac{\kappa'}{\kappa^2 \tau} + B \cdot (\alpha - c) = 0, \tag{17}$
whence
$B \cdot (\alpha - c) = \dfrac{\kappa'}{\kappa^2 \tau} = -\left ( \dfrac{1}{\kappa} \right )' \dfrac{1}{\tau}; \tag {18}$
(5), (10) and (18) express the components of the radial vector $\alpha - c$ in the orthonormal frame $T$, $N$, $B$, whence
$\alpha - c = -\dfrac{1}{\kappa}N -\left ( \dfrac{1}{\kappa} \right )' \dfrac{1}{\tau} B, \tag{19}$
the requisite result.
We may find $r$ in terms of $\kappa$ and $\tau$ by inserting (19) into (1):
$r^2 = (\alpha - c) \cdot (\alpha - c)$
$= \left (-\dfrac{1}{\kappa}N -\left ( \dfrac{1}{\kappa} \right )' \dfrac{1}{\tau} B \right ) \cdot \left (-\dfrac{1}{\kappa}N -\left ( \dfrac{1}{\kappa} \right )' \dfrac{1}{\tau} B \right ); \tag{20}$
since
$\Vert N \Vert = \Vert B \Vert = 1, \; N \cdot B = 0, \tag{21}$
(20) reduces to
$r^2 = \dfrac{1}{\kappa^2} + \left ( \left ( \dfrac{1}{\kappa} \right )' \dfrac{1}{\tau} \right )^2 = \dfrac{1}{\kappa^2} + \left ( -\dfrac{\kappa'}{\kappa^2} \right )^2 \dfrac{1}{\tau^2} = \dfrac{1}{\kappa^2} + \dfrac{(\kappa')^2}{\kappa^4 \tau^2} = \dfrac{(\kappa')^2 + \kappa^2 \tau^2}{\kappa^4 \tau^2}, \tag{22}$
whence
$r = \sqrt{\dfrac{(\kappa')^2 + \kappa^2 \tau^2}{\kappa^4 \tau^2}} = \dfrac{1}{\kappa^2 \tau} \sqrt{(\kappa')^2 + \kappa^2 \tau^2}, \tag{23}$
as the desired formula for $r$.
Best Answer
The Serret-Frenet formula in the case where the curve is not parametrised by arc-length are $$\frac{1}{V_r(t)}\frac{d}{dt}\begin{pmatrix} T\\ N\\ B \end{pmatrix} = \begin{pmatrix} 0 & \kappa & 0\\-\kappa & 0 & \tau \\ 0 &-\tau & 0\end{pmatrix} \begin{pmatrix} T\\N \\B \end{pmatrix}$$ where $V_r(t)=\Vert \dot{r}(t) \Vert$ is the speed of your curve. This is true because in the case of non-arc-legnth parametrisation, the definition of torsion and curvature changes, in order to take into account the speed. Namely, we have that for a $C^3$ biregular curve, the torsion is defined by $$\tau = \frac{1}{V_r} \langle B, \dot{N} \rangle$$ and the curvature $\kappa$ is defined as the norm of the curvature vector $$K = \frac{1}{V_r} \dot{T}$$
Assuming that all these quantities are well defined (which requires, as I mentionned earlier, that the curve be at least $C^3$ and biregular), then it holds that the "trace", or the image of the curve, is invariant under reparametrisation (actually we even need a lot less: assuming merely that the curve is $C^1$ and regular is enough, for instance).
Serret-Frenet formulas provide a cinematic interpretation of the curvature and the torsion. Locally, these quantities tell you how the curve is going to evolve. More generally, it is possible to show that (under the usual hypothesis $C^3$ and biregular) if two curves parametrised by arc-length possess same curvature and same torsion, then they are "equal" in the sense that there exists a deplacement of $\mathbb{R}^3$ from one onto the other. Therefore curvature and torsion control all the geometry of the curve, and thus solving Serret-Frenet completely determines the curve.
A further refinement of this theorem would be that if you arbitrarily give two (continuous) functions, one for curvature, and one for torsion, then there exists a unique curve parametrised by arc-length whose curvature and torsion are the one prescribed (up to a deplacement of the space).