Hints
A. Looking at Figure 01
A cycloid is a curve traced by a point $\:\mathrm P\:$ on the rim of a circular wheel as the wheel rolls along a straight line without slipping.(Wikipedia)
A brachistochrone curve$\rm 1$ or curve of fastest descent, is the one lying on plane between a point $\:\mathrm A\:$ and a lower point $\:\mathrm F$, where $\:\mathrm F\:$ is not directly below A, on which a bead $\:\mathrm P\:$ slides frictionlessly under the influence of a uniform gravitational $\:\mathbf{g}\:$ field in the shortest time.(Wikipedia)
To determine the radius of curvature $\:\rho\:$ of the cycloid using centripetal acceleration we note at first that for any curvilinear motion (not necessarily circular)
\begin{equation}
\Vert\mathbf{a}_{\textrm{n}}\Vert=\dfrac{\Vert\boldsymbol{\upsilon}\Vert^{2}}{\rho}=\dfrac{\upsilon^{2}}{\rho}
\tag{01}
\end{equation}
where $\:\mathbf{a}_{\textrm{n}}\:$ the centripetal acceleration, normal to the curve and $\:\boldsymbol{\upsilon}\:$ the velocity, tangent to the curve.
B. Looking at Figure 02
From the equilibrium of the bead on the direction normal to the cycloid you could determine the magnitude $\:\Vert\mathbf{a}_{\textrm{n}}\Vert\:$ of the centripetal acceleration as function of $\:\mathrm g, \theta$
\begin{equation}
\Vert\mathbf{a}_{\textrm{n}}\Vert=\mathrm h(\mathrm g, \theta)
\tag{02}
\end{equation}
In this Figure it seems that the straight line normal to the cycloid at point $\:\mathrm P\:$ passes through the point $\:\mathrm M\equiv$ contact point of the wheel with the straight line on which it is rolling. Why???
From energy conservation of the bead between the starting point $\:\mathrm A\:$ and present point $\:\mathrm P\:$ you could determine the square of the speed $\:\Vert\boldsymbol{\upsilon}\Vert^{2}\:$ as function of $\:R, \mathrm g, \theta$
\begin{equation}
\Vert\boldsymbol{\upsilon}\Vert^{2}=\mathrm f(R,\mathrm g,\theta)
\tag{03}
\end{equation}
Then from (01) you'll determine the radius of curvature $\:\rho\:$ as function of $\:R,\theta$
\begin{equation}
\rho=\dfrac{\:\:\Vert\boldsymbol{\upsilon}\Vert^{2}}{\Vert\mathbf{a}_{\textrm{n}}\Vert}=\dfrac{\upsilon^{2}}{\Vert\mathbf{a}_{\textrm{n}}\Vert}=\rho(R,\theta)
\tag{04}
\end{equation}
After all these : what is the geometric locus of the centers of curvature ?
The geometric locus of the centers of curvature $\:\mathrm P'\:$ of points of a cycloid $\:C\:$ is a cycloid $\:C'\:$ identical to $\:C\:$ but shifted by the vector $\:(-\pi R, -2R)\:$, see Figure 03.
The new cycloid $\:C'\:$ is the envelope of the family of normal to the cycloid $\:C\:$ straight lines, see Figure 04.
$1\:\:$
For more details on the brachistochrone and tautochrone cycloids see my answer therein : What is the position as a function of time for a mass falling down a cycloid curve?
