Um, how would one convert this rectangular equation into polar form:
(x+1)^2 + (y-4)^2 = 17
Put $\displaystyle x=\rho\cos\theta, \ y=\rho\sin\theta$. Then
$\displaystyle (\rho\cos\theta+1)^2+(\rho\sin\theta-4)^2=17$
$\displaystyle \rho^2(\cos^2\theta+\sin^2\theta)+2\rho\cos\theta-8\rho\sin\theta=0$
$\displaystyle \rho^2+2\rho(\cos\theta-4\sin\theta)=0$