# Polar Coordinate System Transform?

What is the fastest way (fewest trigonometric and square root operations) to transform between one radius and angle to that of a polar coordinate system with a different centerpoint? I.e. the polar equivalent of multiplying by the basis matrix in Cartesian coordinates to convert from one coordinate system to another?

E.g. some back of the napkin math - that I'm hoping there is a more efficient way to do - to transform A (with radius r and angle $\theta$) to S where the two centers are on the x axis separated by distance d (for simplicity the unit vectors of the two systems' 0 angles are assigned as (1, 0), and $r_A \leq d$):

$$x_A = r_A\cos \theta_A \\ y_A = r_A\sin \theta_A \\ x_S = d + x_A \\ y_S = y_A$$

$$r_S = \sqrt{x_S^2 + y_S^2} \\ = \sqrt{(d + x_A)^2 + y_A^2} \\ = \sqrt{d^2 + 2dx_A + x_A^2 + y_A^2} \\ = \sqrt{d^2 + 2dr_A\cos \theta_A + r_A^2}$$

$$\theta_S = \arctan \frac{y_S}{x_S} \\ = \arctan \frac{y_A}{d + x_A} \\ = \arctan \frac{r_A\sin \theta_A}{d + r_A\cos \theta_A}$$

• I don't quite understand the math here. What is A and what is S? little a? Is d distance? Explain every variable you use or no one will be able to make sense of this. – Jeff Strom Sep 5 '15 at 5:27
• I agree, I think you need to clearly define all the variables you are using. – Morgan Rodgers Sep 5 '15 at 5:38
• Added more Latecization and defined some of the vars – Justin Olbrantz Sep 5 '15 at 5:58

$$r = r_0cos(\theta-\phi)+\sqrt{a^2-r_0^2sin^2(\theta-\phi)}$$
This is a circle of radius $a$ centered at the polar coordinate $(r_0,ϕ)$, and with $(r, \theta)$ representing any arbitrary point on the circle. If you want to change location of the circle, simply swap in the new $(r_0,ϕ)$ values.
• Hmm, so that's the relationship between r and $\theta$ for the circle. Are there equations other than mine to find the r and $\theta$ from an r and $\theta$ relative to the other circle? – Justin Olbrantz Sep 5 '15 at 6:52
• You mean like using the new center of the circle as the origin and finding the distance of the old $r$ and $\theta$ to them? I would just convert from polar to cartesian and find the distance, and then convert back. PS accept my answer – Jeff Strom Sep 5 '15 at 7:05