= Solution
Substituting $x=r\cos\theta$, $y=r\sin\theta$ into the <circle> equation gives $r^2-2r\cos\theta=0$. Thus its nonzero radial intersection is $r_Q=2\cos\theta$, with the origin also on the <circle>. The <tangent to a circle> at the rightmost point is $x=2$, or $r_R\cos\theta=2$.
Take rays with $-\pi/2<\theta<\pi/2$. The difference of the two radial distances gives the <cissoid of Diocles>:
$$
\boxed{r_P=2\sec\theta-2\cos\theta=\frac{2\sin^2\theta}{\cos\theta}.}
$$
Multiplying by $\cos\theta$ gives $x=2\sin^2\theta=2y^2/(x^2+y^2)$, hence the Cartesian equation $y^2(2-x)=x^3$. A convenient <parametric curve>, with $t=\tan\theta$, is
$$
x=\frac{2t^2}{1+t^2},\qquad y=\frac{2t^3}{1+t^2},\qquad -\infty<t<\infty.
$$
The real <curve> has $0\leq x<2$, a <cusp> at the origin, symmetry about the $x$ axis, and the vertical <asymptote> $x=2$. Near the origin, $x\sim2t^2$ and $y\sim2t^3$, so its cusp has horizontal tangent direction. As $|t|\to\infty$, $x\to2^-$ while $|y|\to\infty$. The circle and curve meet away from the origin at $(1,\pm1)$.
\Image[/past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2007/paper-2-cissoid.png]
{title=The cissoid has a cusp at the origin and approaches the tangent line}
{description=An original plot of the derived parametric formula, with the unit <circle> centred at $(1,0)$ and its <tangent line> $x=2$.}
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