How to translate a circle to new centre?

Question

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?

Answer 1

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 :

• $r=\sqrt{(x_C+R\sin \theta_0)^2+(y_C+R\cos\theta_0)^2}$
• $\theta=\arctan2\left(\frac x y\right)=\arctan2\left(\frac {x_C+R\sin \theta_0}{y_C+R\cos\theta_0}\right)$

Answer 2

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.

#

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.