Converting Cartesian circle to polar form I am trying to convert circle equation from Cartesian to polar coordinates. I know the solution is all over the Internet but what I am looking for is the exact procedure and explanation, not just the solution. If we start off with: 
$(x-a)^{2} + (y-b)^{2} = r^{2}$
and use 
$x=r\cos{\theta}$
$y=r\sin{\theta}$
I got something that doesn't make a lot of sense. Thanks!
 A: The parametrization would be $x=r\cos\theta+a$, $y=r\sin\theta+b$. Since then $(x-a)^2+(y-b)^2=r^2\cos^2\theta+r^2\sin^2\theta=r^2$
We take $x=r\cos\theta+a$, $y=r\sin\theta+b$, instead of $x=r\sin\theta+a$, $y=r\cos\theta+b$, as it produces a circle that is oriented anticlockwise.
A: The true polar form of a de-centered circle is indeed given by
$$(\rho\cos\theta-a)^2+(\rho\sin\theta-b)^2=\rho^2-2\rho(a\cos\theta+b\sin\theta)+a^2+b^2=r^2.$$
You get an explicit expression by solving the quadratic equation in $\rho$, giving
$$\color{green}{\rho=(a\cos\theta+b\sin\theta)\pm\sqrt{(a\cos\theta+b\sin\theta)^2+r^2-a^2-b^2}}$$ (only the positive roots should be considered).
Note that the term $(a\cos\theta+b\sin\theta)^2$ oscillates between $0$ and $a^2+b^2$ so that when $r<\sqrt{a^2+b^2}$, the domain of $\theta$ is limited.
Of course, when $a=b=0$, this reduces to $\rho=r$.

With the trigonometric transform $a\cos\theta+b\sin\theta=d\cos(\theta-\phi)$, where $d^2=a^2+b^2$,
$$\rho=d\cos(\theta-\phi)\pm\sqrt{r^2-d^2\sin^2(\theta-\phi)}.$$
