# polar representation

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• April 24th 2011, 10:55 AM
frankinaround
polar representation
Hi theres this problem : give the polar representation for the circle of radius 3 centered at (0,3).

I got the equation of the circle : x^2+(y-3)^2=9
Also I know how to change a single point into polar coords. Basicly just draw a triangle and do trig. but how do you turn a equation into polar? Im stumped.
• April 24th 2011, 11:02 AM
TKHunny
$x = r\cdot\cos(\theta)$
$y = r\cdot\sin(\theta)$

Seriously, that's it.
• April 24th 2011, 12:24 PM
sedam7
Quote:

Originally Posted by frankinaround
Hi theres this problem : give the polar representation for the circle of radius 3 centered at (0,3).

I got the equation of the circle : x^2+(y-3)^2=9
Also I know how to change a single point into polar coords. Basicly just draw a triangle and do trig. but how do you turn a equation into polar? Im stumped.

or you can do it like this :D

x = r\cdot\cos(\theta)
y-3 = r\cdot\sin(\theta)
• April 24th 2011, 01:20 PM
frankinaround
well I get that y should be 3Sin(theta) and x should be 3cos(theta). I guess that any point that that works out for will be on the circle for either. but where is the origin in this equation ? Also the answer is supposed to be R=6sin(theta) so im like huh?
• April 24th 2011, 03:29 PM
topsquark
Quote:

Originally Posted by frankinaround
well I get that y should be 3Sin(theta) and x should be 3cos(theta). I guess that any point that that works out for will be on the circle for either. but where is the origin in this equation ? Also the answer is supposed to be R=6sin(theta) so im like huh?

Given an arbitrary point (x, y) extend a line from the origin to the point (x, y). Call the length of this line r and the angle it makes with the +x axis (theta). Using basic trigonometry you can find that x = r*cos(theta) and y = r*sin(theta). Note that the angle (theta) must be measured from the +x axis and not the -x axis.

@ sedam7 y-3 = r*sin(theta) can be used but you must bear in mind that your equation gives a vertical translation of the original y axis. The substitution y = r*sin(theta) is more appropriate here.

As to the rest:

http://latex.codecogs.com/png.latex?x^2 + (y - 3)^2 = 9

http://latex.codecogs.com/png.latex?...ta) - 3)^2 = 9

http://latex.codecogs.com/png.latex?...theta) + 9 = 9

Now simplify. Note that
http://latex.codecogs.com/png.latex?...sin^2(\theta)).

-Dan