Following the carifying comments above, let's go with the second definition. The idea there is that the curve's a circular helix if there's an axis ($P + td$) such that the projection of the curve onto the plane perpendicular to the axis traverses a circle of radius $A$ with speed $AB$ in that plane. That seems to capture circular helix-ness.
Now suppose you have a curve with constant curvature and torsion. To prove it matches the definition, you need to come up with the axis and the constants $A$ and $B$. Well, if you think of a model circular helix around the $z$-axis, say, the normal vector always lies in the $xy$-plane. So you could "discover" the axis by taking any two nearby normals and computing their cross-product. Of course, the more nearby they are, the shorter the cross-product vector will be, so to get a nice unit vector for defining the axis, I'm implicitly suggesting that you say this:
Let $$\vec{d} = \lim_{h \to 0} \frac{1}{h} N(h) \times N(0)$$.
You can rewrite that as
\begin{align}
\vec{d} &= \lim_{h \to 0} \frac{1}{h} (N(h) - N(0)) \times N(0) \\
&=
\left( \lim_{h \to 0} \frac{1}{h} (N(h) - N(0)) \right)\times N(0) \\
&= N'(0) \times N(0)
\end{align}
Now you can use the Frenet-Serret formulas to work out an explicit value for the vector $\vec{d}$. (Note: The result won't generally be a unit vector, but it'll have nonzero length, and then you can make it a unit vector; without the division by $h$, you'd end up with a zero vector.)
Going back to the model of a circular helix around the $z$-axis, you should be able to infer the radius of the circle from the constant curvature and torsion. (Should be 1/curvature, I believe, but I'd have to check whether torsion somehow sneaks in there). That lets you pick a point $P$ at which to start the axis:
$$
P = \gamma(0) + r N(0)
$$
where $\gamma$ is your curve, and $r$ is the radius you computed. For vectors $\vec{e}$ and $\vec{f}$, I'd recommend $N(0)$ and $\vec{d} \times N(0)$. Now all you have to do is show that everything matches up as needed.
Since you didn't ask for a complete proof, but instead for a definition of circular helix that would allow you to prove that constant-curvature-and-torsion curves were indeed circular helices, I think I've given you what you asked for. I can fill in details if necessary, but that might take away the fun from you.
Following comments and discussion
Here's a more complete answer that depends only on the curve $\alpha$, and its Frenet frame. Since comments reveal that you're teaching (or taught) a DG course, I'll let you do the computations here, and just post the main results.
- The helical curve
$$
H(s) = (r \cos us, r \sin us, v)
$$
has tangent vector
$$
T(s) = (-ru \sin us, ru \cos us, v)
$$
which is a unit vector only if
$$
r^2 u^2 + v^2 = 1.
$$
So I'll consider only triples $(r, u, v)$ that satisfy that condition. I'll also restrict to $r > 0$.
The derivative of $T$ is
$$
T'(s) = (-ru^2 \cos us, -ru^2 \sin us, 0)
$$
whose length is $ru^2$; by the Frenet formulas, this is the curvature, $\kappa$, of the helix, and the normal vector to the helix is
$$
N(s) = (-\cos us , \sin us, 0).
$$
The derivative of the normal is
$$
N'(s) = u(\sin us, -\cos us, 0).
$$
The binormal is
$$
B(s) = T \times B = (v \sin ut, -v \cos ut, ru).
$$
The torsion, by Frenet, is $\tau = N' \cdot B = uv$.
So for such helical curves, we have
\begin{align}
ru^2 &= \kappa \\
r^2u^2 + v^2 &= 1 \\
uv &= \tau.
\end{align}
We can solve these for $r, u, v$ to get
\begin{align}
u &= \sqrt{\kappa^2 + \tau^2}\\
v &= \frac{\tau}{\sqrt{\kappa^2 + \tau^2}} \\
r &= \frac{\kappa}{\kappa^2 + \tau^2}
\end{align}
Thus for any positive $\kappa$ and any $\tau$, we can find a unit-speed helical curve $s \mapsto H_{\kappa, \tau}(s)$ by using the values of $u, v, r$ above in $H$.
Now let's look at the constant-speed, constant torsion curve $\alpha$. Let $Q = \alpha(0)$, and let $u, v, r$ be derived from the known curvature and torsion as above.
$$
\newcommand{bT}{\mathbf T}
\newcommand{bN}{\mathbf N}
\newcommand{bB}{\mathbf B}
\newcommand{Tb}{ T_\beta}
\newcommand{Nb}{ N_\beta}
\newcommand{Bb}{ B_\beta}
\newcommand{bD}{\mathbf D}
\newcommand{bE}{\mathbf E}
$$
Let $\bT, \bN, \bB$ denote the unit tangent, normal, and binormal of $\alpha$ at $s = 0$.
Define
\begin{align}
\beta(s) &= (Q + r\bN) - r\cos(us)\bN + r\sin(us) \bE + vs \bD, \text{ where}\\
\bD &= \frac{1}{\|\bN \times N'(0)\|} \bN \times N'(0)\\
&= \frac{1}{\|\bN \times (-\kappa\bT + \tau \bB)\|} \bN \times (-\kappa\bT + \tau \bB)\\
&= \frac{1}{\|\kappa\bB + \tau \bT)\|} (\kappa\bB + \tau \bT)\\
&= \frac{1}{\sqrt{\kappa^2+ \tau^2}} (\kappa\bB + \tau \bT), \text{ and}\\
\bE = \bN \times \bD,
\end{align}
which you can work out in terms of $\bT, \bN,$ and $\bB$ similarly.
Then direct computation shows that
$\beta$ and $\alpha$ start at the same point.
$\beta$ has constant curvature $\kappa$.
3, $\beta$ has constant torsion $\tau$.
By the fundamental theorem of curves, $\beta$ must equal $\alpha$. Bur $\beta$ is evidently a helix with centerline $\bD$. (Indeed, it's the the result of rotating the "standard helix" $H(\kappa, \tau)$ by the matrix whose columns are $\bT, \bN, \bB$, and then translating by $(Q + r\bN)$.)
Best Answer
A cylindrical helix is a curve on a generalized cylinder (take any space curve and form a cylinder by taking parallel lines through each point of the space curve) that makes a constant angle with the rulings (the parallel lines). A circular helix is such a curve on a right circular cylinder. Any such is congruent to a standard helix: $$\alpha(t)=(a\cos t, a \sin t, bt)$$ for some $a>0$, $b\ne 0$.