I am trying to find the equation (in polar form) of a cardioid shifted by an amount $x_0$ in the $x$ direction.
I tried converting $r=1+\cos(\theta)$ in Cartesian coordinates, add $x_0$ to the $x$ component, and convert back to polar. However the new $\theta$ no longer corresponds to the polar angle, so I don't know how to get a closed form $r=f(\theta)$.
Best Answer
Polar equations live poorly with translations.
From $\rho=f(\theta)$ you draw the Cartesian coordinates $(\rho\cos\theta,\rho\sin\theta)$, then after translation $(\rho,\cos\theta+x_0,\rho\sin\theta)$. Reverting to polar,
$$\rho'=\sqrt{(\rho\cos\theta+x_0)^2+(\rho\sin\theta)^2}=\sqrt{\rho^2+2\rho\cos\theta+x_0^2}$$
and
$$\tan\theta'=\frac{\rho\sin\theta}{\rho\cos\theta+x_0}.$$
In general it is not possible to invert the relation between $\theta$ and $\theta'$ and you must content yourself with polar parametric equations.