Conclude that the trace of $\gamma$ is contained in a circle centered in $p$. Let $\gamma$ is a unit speed plane curve with a constant nonzero curvature $k$. 
(a) Prove that the curve $\beta$ defined by $\beta(s)=\gamma(s)+\frac{1}{k^2}\gamma''(s)$ is a constant curve.
(b) Conclude that the trace of $\gamma$ is contained in a circle centered in $p$.
I have done the first part but unable to do the 2nd one.
 A: Given that $\gamma(s)$ is a unit speed plane curve, we have (as is well-known) that the parameter $s$ is the arc-length along $\gamma(s)$, and that the unit tangent vector $T(s)$ to $\gamma(s)$ is given by
$T(s) = \gamma'(s); \tag 1$
we also have the constant curvature $k \ne 0$ (hence in fact $k > 0$) and the unit normal vector $N(s)$ to $\gamma(s)$ satisfy
$\gamma''(s) = T'(s) = kN(s); \tag 2$
thus the curve
$\beta(s) = \gamma(s) + \dfrac{1}{k^2} \gamma''(s) \tag 3$
may, with the aid of (2), be written
$\beta(s) = \gamma(s) + \dfrac{1}{k^2}\gamma''(s) = \gamma(s) + \dfrac{1}{k^2}kN(s) = \gamma(s) + \dfrac{1}{k}N(s); \tag 4$
we differentiate (4) with respect to $s$, using the fact that $k$ is constant:
$\beta'(s) = \gamma'(s) + \dfrac{1}{k}N'(s) = T(s) + \dfrac{1}{k}N'(s); \tag 5$
in addition to (2), we have the other Frenet-Serret equation for plane curves:
$N'(s) = -kT(s); \tag 6$
substituting this in (5) yields
$\beta'(s) = T(s) - \dfrac{1}{k}kT(s) = T(s) - T(s) = 0, \tag  7$
which shows that
$\beta(s) = p, \; \text{a constant}, \tag 8$
answering part (a) of the question.  As for part (b), we may now write (4) as
$\gamma(s) + \dfrac{1}{k}N(s) = \beta(s) = p, \tag 9$
or
$\gamma(s) - p = -\dfrac{1}{k}N(s), \tag{10}$
whence
$\Vert \gamma(s) - p \Vert^2 = (\gamma(s) -p) \cdot (\gamma(s)- p)$
$= (-\dfrac{1}{k}N(s)) \cdot (-\dfrac{1}{k}N(s)) = \dfrac{1}{k^2} N(s) \cdot N(s) = \dfrac{1}{k^2}, \tag{11}$
since
$N(s) \cdot N(s) = 1. \tag{12}$
(11) implies
$\Vert \gamma(s) - p \Vert = \dfrac{1}{k}, \tag{13}$
which affirms that every point of $\gamma(s)$ lies a constant distance $k^{-1}$ from $p$, i.e., in the circle of radius $k^{-1}$ centered at $p$; hence the conclusion required in part (b) is had.
