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/
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