Polar Equation of a Circle

Date: 04/12/97 at 16:26:52
From: Frank P.
Subject: Polar Equation of a Circle

I was wondering if you could provide for me a description of the 
following polar equations for a circle.

   r = a*sin (theta)  and r = a*cos (theta).

What influence does a have on the graph?  What influence does theta 
have on the graph?  What's the difference in using sin or cos, and 
how do they change the graph?

Date: 04/12/97 at 17:34:07
From: Doctor Mitteldorf
Subject: Re: Polar Equation of a Circle

Dear Frank,

You are familiar with writing an equation in x and y and then 
plotting the curve that represents all points obeying that equation.  
Well, you can do the same thing using r and theta instead of x and y.  

In fact, r and theta are related to x and y by

   x = r*cos(theta) and y = r*sin(theta).

Equivalently,

   r^2 = x^2 + y^2 and tan(theta) = y/x.

Another way to write this second equation is 

   sin(theta) = y/sqrt(x^2+y^2).

The obvious way to write the equation for a circle in polar 
coordinates is r = a.  Then theta can be anything.  This makes a 
circle of radius a around the origin as center.  

The equation you've written,  r = a sin(theta), is also a circle, but 
it's not so obvious, and it's not centered about the origin.  It's 
centered around another point.  In a few minutes, you should be 
prepared to answer what that point is, and what the radius of the 
circle is.

>what influence does a have on the graph?

We should be able to answer this in a minute

>what influence does theta have on the graph?

This isn't a meaningful question.  Theta is one of the variables in 
the equation, like x and y were in the cartesian equation.

To see that you have a circle, and to see where the center and the 
radius are, just substitute into your equation r=a sin(theta).  

Let 

   r = sqrt(x^2+y^2)               and let 
   sin(theta) = y/sqrt(x^2+y^2).  

Rearrange the equation and you'll have the form

   x^2 + (y-a/2)^2 = (a/2)^2.

When you see the equation in that form, it should be clear what the 
radius is and where the center is.

Let me know if this is not clear, please.

-Doctor Mitteldorf,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/