Equivalently, let $\gamma(t)$ is a unit-speed curve with $\kappa(t) > 0$ and $\tau(t)$ never equal to zero. We claim $\gamma(t)$ is spherical iff
$$ \frac{\tau}{\kappa} = \frac{d}{ds}\big( \frac{\dot{\kappa}}{\tau\kappa^{2}}\big)$$
First assume that $\gamma(t)$ of spherical. Then there is some point $a$, the center of the sphere, such that
$$||\gamma - a||^2 = r^{2}$$
for the radius of the sphere $r$. Rewrite this as
$$(*) \,\,\,\,\,\,(\gamma - a) \cdot (\gamma - a) = r^{2} $$
To get the equality above, we are going to keep on taking messier and messier derivatives. C'est la vie. You should recall important equalities for unit-speed curves, like $\dot{t} = \kappa n$ and $\dot{b} = -\tau n$. I will use these implicitly in the calculation.
To start, take the derivative of $(*)$ to get
$$ t \cdot (\gamma - a) = 0$$
Now, take the derivative again
$$t \cdot t + \kappa n \cdot (\gamma - a) = 0$$
Note that $t \cdot t = 1$, so we can rewrite this as
$$ n \cdot (\gamma - a) = -\frac{1}{\kappa}$$
Take the derivative of both sides again, to get
$$ n \cdot t + (-\kappa t + \tau b) \cdot (\gamma - a) = \frac{\dot{\kappa}}{\kappa^{2}}$$
Since $t \cdot (\gamma - a) = 0$, this reduces to
$$ b \cdot (\gamma - a) = \frac{\dot{\kappa}}{\tau\kappa^{2}}$$
Take the derivative again to get
$$b \cdot t - \tau n \cdot (\gamma - a) = \frac{d}{ds}\big( \frac{\dot{\kappa}}{\tau\kappa^{2}}\big)$$
which is easily seen to be equivalent to
$$ \frac{\tau}{\kappa} = \frac{d}{ds}\big( \frac{\dot{\kappa}}{\tau\kappa^{2}}\big)$$
Now, we go the other direction, so assume
$$(*) \,\,\,\,\,\, \frac{\tau}{\kappa} = \frac{d}{ds}\big( \frac{\dot{\kappa}}{\tau\kappa^{2}}\big)$$
Let us define new quantities $\rho = 1/\kappa$ and $\sigma = 1/\tau$. Then $(*)$ implies that
$$(\dagger) \,\,\,\,\, \rho = \frac{d}{ds}\big(-\sigma(\dot{\rho}\sigma)\big)$$
Consider the quantity $\rho^2 + (\dot{\rho}\sigma)^2$. Let us take its derivative. By $(\dagger)$, we would have
$$\frac{d}{ds} (\rho^2 + (\dot{\rho}\sigma)^2) = 2\rho\dot{\rho} + 2(\rho\sigma)\frac{d}{ds}(\rho\sigma) = 0$$
Hence $\rho^2 + (\dot{\rho}\sigma)^2$ is some positive constant $r^2$. Let us go further and define the following "curve" $a$ (think about its definition a little)
$$a = \gamma + \rho n + \dot{\rho}\sigma b$$
Now, we take the derivative of $a(t)$, noting the equation following $(\dagger)$,
$$\dot{a} = t + \dot{\rho}n + \rho(-\kappa t + \tau b) + \frac{d}{ds} (\dot{\rho}\sigma)b + (\dot{\rho}\sigma)(-\tau n) = 0$$
and so $a$ is a constant, and is the center of our sphere by the following calculation
$$\parallel \gamma - a \parallel^2 = ||\rho n + \dot{\rho}\sigma b||^2 = \rho^2 + (\dot{\rho}\sigma)^2 = r^2$$
and the radius of this sphere is $r$. This demonstrates that $\gamma(t)$ lies on the surface of a sphere of radius $r$, and hence is spherical. QED.
thanks to the coments of @willjagy i could conclude my demonstrations, first i find by derivating $|\alpha(s)|^2$ that $<T,\alpha>=0$, derivating again and using Frenet Formulas i get
$k<N,\alpha>+1=0$ and one more time:
$k^{'}<N,\alpha>-k\tau<B,\alpha>=0$, where $\tau$ is the torsion of the curve, so, as $k$ is contant and positive:
$\tau<B,\alpha>=0$ $ \forall s\in I$,
and then at any cases $\tau=0$ or $<B,\alpha>=0$ i can conclude the $\alpha$ is a circle, for the first i get $\alpha$ in a great circle of the sphere and for the second an arbitrary cirlce with radius less or equal than $r$.
Best Answer
Let's denote the Frenet frame of $\alpha$ by $\{T,N,B\}$. As you say, let's differentiate the equality $$ (\alpha-p)\cdot(\alpha-p)=a^{2}, $$ which gives $$ T\cdot\alpha=0.\tag{1} $$ Differentiating once more and using the Frenet equations, we get $$ kN\cdot\alpha+1=0 \Rightarrow r=\frac{1}{k}=-N\cdot\alpha.\tag{2} $$ Then $$ r'=-N'\cdot\alpha-N\cdot T=(kT-\tau_{0}B)\cdot\alpha=-\tau_{0}B\cdot\alpha, $$ using (1) in the last equality. Differentiate once more: $$ r''=-\tau_{0}(-\tau_{0}N\cdot\alpha+B\cdot T)=\tau_{0}^{2}N\cdot\alpha=-\tau_{0}^{2}r, $$ using (2) in the last equality.