# Converting polar equation to cartesian coordinate polar equation and back again?

OK, so I have the following polar equation:

$r = Θ/20$

And I would like to translate this a little to the right, and down from the polar origin.

Now, I figure since I know cartesian coordinate equation translations are quite simple, the best way to do this would be to convert this polar equation to a rectangular equation, translate, and then convert back to polar again.

However, I was having some problems making the initial conversion.

I know we are supposed to use the following relationships between polar and rectangular equations:

$r^2 = x^2 + y^2$

$y = rsin(Θ)$

But, I cannot seem to convert my equation correctly...

Would someone be able to suggest a simpler way to make this translation?

Or, assist me in converting to polar and back again?

I'm quite new at this, so I apologize if this question is remedial.

Thanks!

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The other equation you need is $\theta = \arctan \frac {y}{x}$ and you have to worry about the branches of arctan. So now you have $\sqrt{x^2+y^2}=\frac{1}{20} \arctan \frac {y}{x}$. To translate right and down, you replace $x$ with $x+a$ and $y$ with $y-b$. And you have quite a mess. Taking $\sqrt{(x+a)^2+(y-b)^2}=\frac{1}{20} \arctan \frac {y-b}{x+a}$ back to polar is a lot of work and probably not illuminating.
If you want to do a translation in polar coordinates by $(r_0,\theta_0)$, the following formulas hold $$r' = \sqrt{r^2 + r_0^2 -2 r r_0 \cos(\theta - \theta_0)}$$ $$\theta' = \arctan([ r \sin(\theta) -r_0 \sin(\theta_0)]/[r \cos(\theta) -r_0 \cos(\theta_0)])$$ where $'$ denotes the new system. Note that there is a problem with the ambiguity of the $\arctan$ function. For this reason often arctan2 is implemented (or you can implement it yourself). Having arctan2 the second equation reads $$\theta' = \text{arctan2}( r \sin(\theta) -r_0 \sin(\theta_0),r \cos(\theta) -r_0 \cos(\theta_0)).$$