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Call the centre of the circle $O$, the left endpoint on the diameter $A$, the point of tangency $T$, and the external intersection $E$. Since $OA=OT$ are radii, the isosceles triangle $AOT$ has equal base angles: $\angle OAT=\angle ATO=30^{\circ }$. Therefore $\angle AOT=120^{\circ }$, and, because $OA$ and $OE$ point in opposite directions, $\angle TOE=60^{\circ }$. The tangent to a circle $TE$ is perpendicular to the radius $OT$, so in the right triangle $OTE$,

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: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 $t=\tan \theta$, isThe real curve has $0\le x<2$, a cusp at the origin, symmetry about the $x$ axis, and the vertical asymptote $x=2$. Near the origin, $x\sim 2t^2$ and $y\sim 2t^3$, so its cusp has horizontal tangent direction. As $|t|\to \infty$, $x\to 2^{-}$ while $|y|\to \infty$. The circle and curve meet away from the origin at $(1,\pm 1)$.