Let $\gamma$ be a parameterization by arc length of a curve $C$. Prove that,
if there are: $X_0 \in \mathbb{R}^3$, and $\alpha,\beta \in [a,b] \to \mathbb{R}$, differentiable on $[a, b]$, such that $\alpha(s)N(s) + \beta(s)B(s) +\gamma(s) = X_0$ for all $s \in [a,b]$, then $C$ lies on a sphere.
Given that
$X_0 = \gamma(s) + \alpha(s)N(s) + \beta(s) B(s), \tag 1$
we have
$X_0 - \gamma(s) = \alpha(s)N(s) + \beta(s) B(s), \tag 2$
whence
$\Vert X_0 - \gamma(s) \Vert ^2 = (X_0 - \gamma(s)) \cdot X_0 - \gamma(s))$ $= (\alpha(s)N(s) + \beta(s) B(s)) \cdot (\alpha(s)N(s) + \beta(s) B(s)) = \alpha^2(s) + \beta^2(s), \tag 3$
since
$N(s) \cdot N(s) = 1 = B(s) \cdot B(s) \tag 4$
and
$N(s) \cdot B(s) = 0. \tag 5$
We next differentiate (1) with respect to $s$, observing that $X_0 \in \Bbb R^3$ is fixed:
$0 = \dfrac{d}{ds}(X_0)$ $= \dot{\gamma}(s) + \dot{\alpha}(s)N(s) + \alpha(s) \dot N(s) + \dot{\beta}(s) B(s) + \beta(s) \dot B(s), \tag 6$
and recalling that
$\dot \gamma(s) = T(s), \tag 7$
and the Frenet-Serret equations
$\dot N(s) = -\kappa(s) T(s) + \tau(s) B(s), \tag 8$
$\dot B(s) = -\tau(s) N(s), \tag 9$
we transform (6) into
$0 = \dfrac{d}{ds}(X_0)$ $= T(s) + \dot{\alpha}(s)N(s) + \alpha(s) (-\kappa(s) T(s) + \tau(s) B(s)) + \dot{\beta}(s) B(s) - \beta(s) \tau(s) N(s)$ $ = T(s) + \dot{\alpha}(s)N(s) - \alpha(s) \kappa(s) T(s) + \alpha(s) \tau(s) B(s) + \dot{\beta}(s) B(s) - \beta(s) \tau(s) N(s)$ $= (1 - \alpha(s) \kappa(s)) T(s) + (\dot \alpha(s) - \beta(s) \tau(s))N(s) + (\dot \beta(s) + \alpha(s) \tau(s))B(s); \tag {10}$
by virtue of the orthogonality of $T(s)$, $N(s)$ and $B(s)$ we may infer that
$\alpha(s) \kappa(s) = 1, \tag{11}$
$\dot \alpha(s) = \beta(s) \tau(s), \tag{12}$
$\dot \beta(s) = -\alpha(s) \tau(s); \tag{13}$
thus,
$\alpha(s) \dot \alpha(a) + \beta(s) \dot \beta(s) = \alpha(s) \beta(s) \tau(s) - \beta(s) \alpha(s) \tau(s) = 0, \tag{14}$
which is easily seen to imply that
$\dfrac{d}{ds}(\alpha^2(s) + \beta^2(s)) = 0, \tag{15}$
and we conclude that $\alpha^2(s) + \beta^2(s)$ is constant; returning to (3), we see that this implies that $\gamma(s)$ lies in the sphere of radius $\sqrt{\alpha^2(s) + \beta^2(s)}$ centered at $X_0$.
A Few Further Remarks: First, we observe that (11) may be written
$\alpha(s) = \dfrac{1}{\kappa(s)}, \tag{16}$
which shows not only that
$\alpha(s) > 0, \tag{17}$
but also that $\alpha(s)$ is the radius curvature of $\gamma(s)$ at $s$. Second, (16) yields
$\dot \alpha(s) = -\dfrac{\dot \kappa(s)}{\kappa^2(s)}, \tag{18}$
and combining this with (12) we find
$\beta(s) \tau(s) = -\dfrac{\dot \kappa(s)}{\kappa^2(s)}, \tag{19}$
or, in the event $\tau(s) \ne 0$,
$\beta(s) = -\dfrac{\dot \kappa(s)}{\kappa^2(s) \tau(s)} = \dfrac{d}{ds} \left ( \dfrac{1}{\kappa(s)} \right )\left ( \dfrac{1}{\tau(s)} \right ). \tag{20}$
In light of (16) and (20) we see that (2) may be expressed in the form
$X_0 - \gamma(s) = \dfrac{1}{\kappa(s)} N(s) - \dfrac{\dot \kappa(s)}{\kappa^2(s) \tau(s)} B(s)$ $= \dfrac{1}{\kappa(s)} N(s) + \dfrac{d}{ds} \left ( \dfrac{1}{\kappa(s)} \right )\left ( \dfrac{1}{\tau(s)} \right )B(s), \tag{21}$
which in accord with (3) shows that the radius of the sphere, $\Vert \gamma(s) - X_0 \Vert$, satisfies
$\Vert \gamma(s) - X_0 \Vert^2 = \dfrac{1}{\kappa^2(s)} + \dfrac{\dot \kappa^2(s)}{\kappa^4(s) \tau^2(s)}. \tag{22}$
This formula has appeared elsewhere on this site, see for example this question.
From (20),
$\dot \beta(s) = -\dfrac{\ddot \kappa(s)(\kappa^2(s) \tau(s)) - \dot \kappa(s) (\kappa^2(s) \tau(s))'}{\kappa^4(s) \tau^2(s)}$ $= -\dfrac{\ddot \kappa(s)(\kappa^2(s) \tau(s)) - \dot \kappa(s) (2\kappa(s) \dot \kappa(s) \tau(s) + \kappa^2(s) \dot \tau(s))}{\kappa^4(s) \tau^2(s)}$ $= -\dfrac{\ddot \kappa(s)(\kappa^2(s) \tau(s)) - 2\kappa(s) \dot \kappa^2(s) \tau(s) - \kappa^2(s) \dot \kappa(s) \dot \tau(s)}{\kappa^4(s) \tau^2(s)}. \tag{23}$
We note that (11) implies that
$\alpha(s) > 0; \tag{24}$
if
$\beta(s) = 0, \tag{25}$
then (12) yields
$\dot \alpha(s) = 0, \tag{26}$
that is, $\alpha(s)$ is a positive real constant; in light of (25) we also have
$\dot \beta(s) = 0, \tag{27}$
and then via (13) we see that
$\tau(s) = 0, \tag{28}$
which is sufficient that $\gamma(s)$ lie in a plane (see my answer to this question, click here); now (2) becomes
$X_0 - \gamma(s) = \alpha(s)N(s), \tag {29}$
from which, in accord with (3),
$\Vert X_0 - \gamma(s) \Vert^2 = \alpha^2(s), \tag{30}$
or
$\Vert X_0 - \gamma(s) \Vert = \alpha(s) = \dfrac{1}{\kappa(s)}, \tag{31}$
whence since $\alpha(s)$ and $\kappa(s)$ are constant, indicates that $\gamma(s)$ lies in the sphere or radius $\alpha(s)$ centered at $X_0$; when combined with the fact that $\gamma(s)$ lies in a plane, allows the conclusion that $\gamma(s)$ in fact lies in a circle of radius $\alpha(s)$ centered at $X_0$.
We have seen above (ca. (28)) that $\gamma(s)$ lies in a plane;
Need to re-save in .txt file!
End of Remarks.
