Describing all plane curves with constant curvature I know that by Frenet-Serret, we have (I know this is only for curves parametrized by arclength, but since every plane curve can be reparametrized by arclength, there's no loss of generality):
$t'(s) = k(s)n(s) \Rightarrow t''(s) = k(s)n'(s)$ (since $k(s)$ is constant)
$n'(s) = -k(s)t(s)$
Since the curvature is constant, I can call $k(s) = c$ and get:
$t''(s) = -c^2 t(s)$
$t''(s) + c^2t(s) = 0$
which has solutions:
$t(s) = A_1 \cos(cs) + A_2 \sin(cs)$ (where I also assume $||t|| = ||n|| = 1$)
so
$a(s) = \frac{1}{c} (A_1 \sin(cs) - A_2 \cos(cs)) + A_3$ is the form all regular curve planes are (I forgot to add regular in the beginning, but it's in the exercise)
"Squaring" both sides, we get:
$||a(s) - A_3||^2 = \frac{1}{c^2} ||A_1||^2 \Rightarrow ||a(s) - A_3||^2 = \frac{1}{c^2} \Rightarrow ||a(s) - A_3|| = \frac{1}{c}$, which is clearly the equation of a circle (I used Gribouillis' answer in the middle of these steps)
 A: You have
$$
1 = \|t\|^2 = (A_1 \cos(cs) + A_2\sin(cs))^2 = A_1^2 \cos^2(cs) + A_2^2 \sin^2(cs) + 2 \cos(cs)\sin(cs)(A_1\cdot A_2)
$$
Differentiating gives (we assume $c\not=0$)
$$
(A_2^2 - A_1^2)2\cos(cs)\sin(cs)+ A_1\cdot A_2 2(\cos^2(cs) - \sin^2(cs)) = 0
$$
which you can rewrite as
$$
(A_2^2 - A_1^2)\sin(2cs)+ 2 A_1\cdot A_2 \cos(2cs) = 0
$$
derivating again gives
$$
(A_2^2 - A_1^2)\cos(2cs)- 2 A_1\cdot A_2 \sin(2cs) = 0
$$
It follows that $A_1^2 = A_2^2$ and $A_1\cdot A_2 = 0$. Inserting this
in the first equation above shows that $A_1$ and $A_2$ are two orthogonal vectors with norm $1$.
With very little work, you can now conclude that the curve is a circular arc.
A: The OP's solution appears essentially correct to Yours Truly, though as suggested by Gribouillis in his comment it might be wise to say a bit more about $A_1, A_2$ and $K_1$, $K_2$, $K_3$.
Having typed these words, I present a more geometrically-flavored solution which does not rely so much on differential equations or their exact solutions.  To wit:
Suppose $\gamma(s)$ is our curve, parametrized by arc-length, and that
$\kappa(s) = 0; \tag 1$
then
$T'(s) = \kappa(s)N(s) = 0, \tag 2$
so
$T(s) = T_0, \tag 3$
a constant.  Then
$\dot \gamma(s) = T(s) = T_0, \tag 4$
whence
$\gamma(s) - \gamma(s_0) = \displaystyle \int_{s_0}^s \dot \gamma(u) \; du = \int_{s_0}^s T_0 du = T_0(s - s_0), \tag 5$
so
$\gamma(s) = T_0 (s - s_0) + \gamma(s_0), \tag 6$
which is the equation of a straight line.  
If, on the other hand, we have constant
$\kappa(s) = \kappa > 0, \tag 7$
we consider the curve $c(s)$ defined by
$c(s) = \gamma(s) + \kappa^{-1} N(s); \tag 8$
we have
$\dot c(s) = \dot \gamma(s) + \kappa^{-1} \dot N(s) = T(s) + \kappa^{-1} (-\kappa T(s)) = T(s) - T(s) = 0; \tag 9$
it thus follows that the "curve" $c(s)$ is constant:
$c(s) = c_0; \tag{10}$
then (8) becomes
$c_0 = \gamma(s) + \kappa^{-1}N(s), \tag{11}$
or
$\gamma(s) - c_0 = -\kappa^{-1}N(s), \tag{12}$
and thus
$\Vert \gamma(s) - c_0 \Vert = \Vert -\kappa^{-1} N(s) \Vert = \kappa^{-1} \Vert N(s) \Vert = \kappa^{-1}, \tag{13}$
since $\Vert N(s) \Vert = 1$.  (13) indicates that every point or $\gamma(s)$ is the same distance $\kappa^{-1}$ from $c_0$; thus it is the equation of a circle, which in fact $\gamma(s)$ is.
We see that every constant curvature planar curve is either a straight line or a circle.
