How to translate a circle to new centre? Consider that we have a circle drawn a round the origin (0,0). That circle has some points drawn on its circumference. Each of those points has range and azimuth $(r,\theta)$, the r and $\theta$ values of these points are calculate with responding to the origin (0,0). 
I want to move or translate that circle and its (on-circumference) points to a new center (x,y), the new center can be in any quadrant. Here is an image for more demonstration (consider that it was translated to the first quadrant):
The question is what is the new $(r, \theta)$ of each point with responding to the origin (0,0) after translation to the new center?
 A: For a circle centered at the origin we have (taking $x$ axis on the right and $y$ axis upward and $\theta$ as indicated in the sketch):
$$x^2+y^2=R^2 \iff (x,y)=(R\sin \theta_0,R\cos\theta_0)$$
for a circle centered at $C=(x_C,y_C)$ we have:
$$(x-x_C)^2+(y-y_C)^2=R^2\iff (x,y)=(x_C+R\sin \theta_0,y_C+R\cos\theta_0)$$
Then the new polar coordinates are (using the arctan2 function) are :


*

*$r=\sqrt{(x_C+R\sin \theta_0)^2+(y_C+R\cos\theta_0)^2}$

*$\theta=\arctan2\left(\frac x y\right)=\arctan2\left(\frac {x_C+R\sin \theta_0}{y_C+R\cos\theta_0}\right)$

A: In general, expressions in polar form after a translation are not very elegant. 
However, here is how you can derive the polar form of a circle that is not centered at the origin, $\mathbf{O}$.
Consider an arbitrary point $\mathbf{P}(r,\theta)$ on the (translated) circle whose [new] center is at $\mathbf{S}(r_0, \theta_0)$.
Then using the cosine rule on triangle $\mathbf{OPS}$, we have 
$$ r^2 + r_0^2 - 2r_0 r \cos(\theta - \theta_0) = R^2 $$
Thus we have an (implicit) formula for the translated circle which involves $r$ and $\theta$.
Rearranging this into descending powers of $r$, gives.
$$ r^2 -  2r_0 \cos(\theta - \theta_0) r  + (r_0^2 -R^2) = 0 $$.
Since this is a quadratic in $r$, and using the identity $\cos^2 \alpha + \sin^2 \alpha =1 $, we can solve for $r$. 
This results in the more direct expression for $r$ in terms of $\theta$:
$$
r = r_0\cos(\theta-\theta_0)+\sqrt{R^2-r_0^2\sin^2(\theta-\theta_0)}
$$
For a diagram of this, see here.
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Note that standard math notation is East + going anticlockwise. The OP's notation is North + clockwise. For this problem due to it all being in polar notation, using either convention will result in the same final expression.
