Moving a Circle on a Polar Graph

Date: 05/04/2001 at 15:40:50
From: Mike
Subject: Moving a circle on the Polar Graph

Dr. Math,

I was wondering how you move a circle so the center is not (0,0), but 
to where the circle can be (r,[theta]). I tried to read about it in 
the archives, but it was confusing with the multiple equations and 
various [theta]1. I was able to figure out how to mave a circle 
around the pole, but that was as far as I got. 

Thank you very much for your help.
Mike

Date: 05/04/2001 at 17:01:21
From: Doctor Peterson
Subject: Re: Moving a circle on the Polar Graph

Hi, Mike.

I suppose you are referring to our FAQ:

  http://mathforum.org/dr.math/faq/formulas/faq.polar.html#circles   

All you have to look at there is the "General form":

    r^2 - 2r(h cos[theta] + k sin[theta]) = R^2 - h^2 - k^2

That's the answer! It's ugly; polar form doesn't tend to deal well 
with translations, but likes everything nicely centered at the origin. 
But anything CAN be written in polar form, even if it's not very 
useful.

But where does this equation come from?

All we have to do is take the rectangular form of the equation and 
replace (x,y) with (r cos(theta), r sin(theta)):

    (x - h)^2 + (y - k)^2 = R^2

becomes

    (r cos(theta) - h)^2 + (r sin(theta) - k)^2 = R^2
or
    r^2 cos^2(theta) - 2rh cos(theta) + h^2 +
    r^2 sin^2(theta) - 2rk sin(theta) + k^2 = R^2

The first term on each line add up to r^2; if we move the h^2 and k^2 
to the right we get

    r^2 - 2rh cos(theta) - 2rk sin(theta0 = R^2 - h^2 - k^2

I'll leave the last step to you.

The explanation of r1 and theta1 just tells you how to change (h,k) 
into polar form if you need to, which you should already know; it's 
just the definition.

If you want to write the polar equation as a function, r = f(theta), 
you will have to solve for r using the quadratic formula. This will 
make it really ugly! But you will want to try deriving the form for a 
circle passing through the center; try setting (h,k) to (R,0) or 
(0,R), for example, and simplifying the equation.

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/