= Solution
The original PDF measures $\theta$ at the rim point $O$, from the positive $x'$-axis to $OP$; it is not the angle at the centre of the circle. The converted TeX omits this entire subpart. In rotating <Cartesian coordinate system>, the centre is $(a,0)$, and the nonzero intersection of the ray $\mathbf r=\rho(\cos\theta,\sin\theta)$ with $(x'-a)^2+y'^2=a^2$ has $\rho=2a\cos\theta$. Therefore
$$
\boxed{\mathbf r=2a\cos\theta(\cos\theta,\sin\theta,0)=a(1+\cos2\theta,\sin2\theta,0).}
$$
The second form stays regular even at the rim point $O$, where the polar radius is zero. Define the outward <unit normal> from the centre and the increasing-angle <unit tangent vector> by
$$
\mathbf n=(\cos2\theta,\sin2\theta,0),\qquad\boldsymbol\tau=(-\sin2\theta,\cos2\theta,0).
$$
The first two <derivatives> in the <rotating reference frame> are $\mathbf v'=2a\dot\theta\boldsymbol\tau$ and $\mathbf a'=2a\ddot\theta\boldsymbol\tau-4a\dot\theta^2\mathbf n$. These describe a <bead on a horizontal hoop rotating about its rim>.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2017/ia/paper-4-hoop-geometry.png]
{title=The geometry of a sliding hoop and its contact point}
{description=Original constructed geometry of the horizontal hoop: theta is at the rim pivot, and the central angle is twice theta.}
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