If the curvature of a curve in a plane is a constant, is the curve contained in a circle? If the curvature of a curve in a plane is a constant, is the curve contained in a circle?(Suppose curvature is positive.)
one of my homework problems needs to use this, but I am not sure whether this conclusion is right or not? I still have no idea how to prove it. Can someone tell me whether it is right? 
How to prove it? Thank you for help.
 A: Here's how I'd prove it:
Let $\alpha(s)$ be a plane curve, parametrized by it's arc-length $s$; then the unit tangent vector $T(s)$ to the curve is given by
$T(s) = \dot \alpha(s), \tag{1}$
and the unit normal vector $N(s)$ satisfies
$\dot T(s) = \kappa(s) N(s); \tag{2}$
(2) is one of the well-known Frenet-Serret equations for $T$ and $N$; its companion equation for $\dot N(s)$ is
$\dot N(s) = -\kappa(s) T(s); \tag{3}$
we accept (2) and (3) as understood by the majority of the readership who I assume have seen them before.  Now, given a curve $\alpha(s)$, and an initial value $s_0$ of $s$, we have the unit tangent and normal vectors $T_0 = T(s_0)$ and $N_0 = N(s_0)$ at $\alpha(s_0)$, and observing that (2)-(3) together comprise a linear system of ordinary differential equations for $T(s)$ and $N(s)$, we consider, whether $\kappa(s)$ is constant or not, the expressions
$T(s) = T_0 \cos \int_{s_0}^s \kappa(u) du - N_0 \sin \int_{s_0}^s \kappa(u) du \tag{4}$
and 
$N(s) = -N_0 \cos \int_{s_0}^s \kappa(u) du - T_0 \sin \int_{s_0}^s \kappa(u) du. \tag{5}$
Differentiating (4) we find
$\dot T(s) = -\kappa(s) T_0 \sin \int_{s_0}^s \kappa(u) du - \kappa(s) N_0 \cos \int_{s_0}^s \kappa(u) du = \kappa(s) N(s), \tag{6}$
and likewise (5) yields
$\dot N(s) = \kappa(s) N_0 \sin \int_{s_0}^s \kappa(u) du -\kappa(s) T_0 \cos \int_{s_0}^s \kappa(u) du = -\kappa(s) T(s). \tag{7}$
(6) and (7) show that (4) and (5) solve (2) and (3); uniquely so, in fact, since all the  requisite conditions hold (recall that linear ODEs always satisfy Lipschitz continuity etc.).  From (4) it follows that
$\dot \alpha(s) = T(s) = T_0 \cos \int_{s_0}^s \kappa(u) du - N_0 \sin \int_{s_0}^s \kappa(u) du. \tag{8}$
For general $\kappa(s)$, integrating (8) to obtain $\alpha(s)$ proves difficult indeed; however, when $\kappa(s) = \kappa$ is a constant, this assessment may be substantially altered.  In this case, we see that
$\int_{s_0}^s \kappa(u)du = \kappa (s - s_0), \tag{9}$
whence
$\dot \alpha(s) = T_0 \cos \kappa (s - s_0) - N_0 \sin \kappa (s - s_0), \tag{10}$
which in turn may be easily integrated:
$\alpha(s) - \alpha(s_0) = \int_{s_0}^s \dot \alpha(u) du = \int_{s_0}^s (T_0 \cos \kappa (u - s_0) - N_0 \sin \kappa (u - s_0)) du$
$= \dfrac{1}{\kappa} (T_0 \sin \kappa (u - s_0) + N_0 \cos (\kappa (u - s_0) \mid_{s_0}^s$
$= \dfrac{1}{\kappa} (T_0 \sin \kappa (s - s_0) + N_0 \cos (\kappa (s - s_0)) - \dfrac{1}{\kappa} N_0; \tag{11}$
(11) may be re-written as
$\alpha(s) - (\alpha(0) - \dfrac{1}{\kappa}N_0) = \dfrac{1}{\kappa} (T_0 \sin \kappa (s - s_0) + N_0 \cos (\kappa (s - s_0)); \tag{12}$
from this we see that
$\Vert \alpha(s) - (\alpha(0) - \dfrac{1}{\kappa}N_0) \Vert^2 = \Vert \dfrac{1}{\kappa} (T_0 \sin \kappa (s - s_0) + N_0 \cos (\kappa (s - s_0)) \Vert^2$
$ = \langle \dfrac{1}{\kappa} (T_0 \sin \kappa (s - s_0) + N_0 \cos (\kappa (s - s_0)), \dfrac{1}{\kappa} (T_0 \sin \kappa (s - s_0) + N_0 \cos (\kappa (s - s_0)) \rangle$
$= \dfrac{1}{\kappa^2} (\langle T_0, T_0 \rangle \sin^2 \kappa (s - s_0) + 2\langle T_0, N_0 \rangle \sin \kappa (s - s_0) \cos \kappa (s - s_0) + \langle N_0, N_0 \rangle \cos^2 \kappa (s - s_0))$
$ = \dfrac{1}{\kappa^2}, \tag{13}$
where we have used the facts
$\langle T_0, T_0 \rangle = \langle N_0, N_0 \rangle = 1, \tag{14}$
$\langle T_0, N_0 \rangle = 0 \tag{15}$
in the evaluation of (13).  When the intermediate, computational steps are bypassed, (13) reads
$\Vert \alpha(s) - (\alpha(0) - \dfrac{1}{\kappa}N_0) \Vert^2 = \dfrac{1}{\kappa^2}, \tag{16}$
which shows that $\alpha(s)$ lies in the circle of radius $1/\kappa$ centered at $\alpha(0) - (1/\kappa) N_0$.  QED.
Hope this helps.  Cheers,
and as ever,
Fiat Lux!!!
A: HINT: If the curve were (an arc of) a circle, how would you find its center, using the arclength parametrization $\alpha(s)$, curvature $\kappa$, and unit tangent and normal?
