Watt's curve

The polar equation for the curve can be derived as follows:[1] Working in the complex plane, let the centers of the circles be at a and −a, and the connecting segment have endpoints at −a+be^i λ and a+be^{i ρ}. Let the angle of inclination of the segment be ψ with its midpoint at re^{i θ}. Then the endpoints are also given by re^{i θ} ± ce^{i ψ}. Setting expressions for the same points equal to each other gives

${\displaystyle a+be^{i\rho }=re^{i\theta }+ce^{i\psi }.\,}$

${\displaystyle -a+be^{i\lambda }=re^{i\theta }-ce^{i\psi }\,}$

Add these and divide by two to get

${\displaystyle re^{i\theta }={\tfrac {b}{2}}(e^{i\rho }+e^{i\lambda })=b\cos({\tfrac {\rho -\lambda }{2}})e^{i{\tfrac {\rho +\lambda }{2}}}.}$

Comparing radii and arguments gives

${\displaystyle r=b\cos \alpha ,\ \theta ={\tfrac {\rho +\lambda }{2}}\ {\mbox{where}}\ \alpha ={\tfrac {\rho -\lambda }{2}}.}$

Similarly, subtracting the first two equations and dividing by 2 gives

${\displaystyle ce^{i\psi }-a={\tfrac {b}{2}}(e^{i\rho }-e^{i\lambda })=ib\sin \alpha e^{i\theta }.}$

Write

${\displaystyle a=a\cos \theta \ e^{i\theta }-ia\sin \theta \ e^{i\theta }.\,}$

Then

${\displaystyle ce^{i\psi }=ib\sin \alpha e^{i\theta }+a\cos \theta \ e^{i\theta }-ia\sin \theta \ e^{i\theta }=(a\cos \theta \ +i(b\sin \alpha -a\sin \theta ))e^{i\theta },}$

${\displaystyle c^{2}=a^{2}\cos ^{2}\theta +(b\sin \alpha -a\sin \theta )^{2},\,}$

${\displaystyle b\sin \alpha =a\sin \theta \pm {\sqrt {c^{2}-a^{2}\cos ^{2}\theta }},\,}$

${\displaystyle r^{2}=b^{2}\cos ^{2}\alpha =b^{2}-b^{2}\sin ^{2}\alpha =b^{2}-\left[a\sin \theta \pm {\sqrt {c^{2}-a^{2}\cos ^{2}\theta }}\right]^{2}.,\,}$

Expanding the polar equation gives

${\displaystyle r^{2}=b^{2}-(a^{2}\sin ^{2}\theta \ +c^{2}-a^{2}\cos ^{2}\theta \pm 2a\sin \theta {\sqrt {c^{2}-a^{2}\cos ^{2}\theta }}),\,}$

${\displaystyle r^{2}-a^{2}-b^{2}+c^{2}+2a^{2}\sin ^{2}\theta =\pm 2a\sin \theta {\sqrt {c^{2}-a^{2}\cos ^{2}\theta }}),\,}$

${\displaystyle (r^{2}-a^{2}-b^{2}+c^{2})^{2}+4a^{2}(r^{2}-a^{2}-b^{2}+c^{2})\sin ^{2}\theta +4a^{4}\sin ^{4}\theta =4a^{2}\sin ^{2}\theta (c^{2}-a^{2}\cos ^{2}\theta ),\,}$

${\displaystyle (r^{2}-a^{2}-b^{2}+c^{2})^{2}+4a^{2}(r^{2}-b^{2})\sin ^{2}\theta =0,\,}$

${\displaystyle (x^{2}+y^{2})(x^{2}+y^{2}-a^{2}-b^{2}+c^{2})^{2}+4a^{2}y^{2}(x^{2}+y^{2}-b^{2})=0.\,}$

Letting d ²=a²+b²–c² simplifies this to

${\displaystyle (x^{2}+y^{2})(x^{2}+y^{2}-d^{2})^{2}+4a^{2}y^{2}(x^{2}+y^{2}-b^{2})=0.\,}$