# Math Help - polar representation

1. ## 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.

2. $x = r\cdot\cos(\theta)$
$y = r\cdot\sin(\theta)$

Seriously, that's it.

3. 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

x = r\cdot\cos(\theta)
y-3 = r\cdot\sin(\theta)

4. 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?

5. 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:

$x^2 + (y - 3)^2 = 9$

$(r~cos(\theta))^2 + ((r~sin(\theta) - 3)^2 = 9$

$r^2~cos^2(\theta) + r^2~sin^2(\theta) - 6r~sin(\theta) + 9 = 9$

Now simplify. Note that
$r^2~cos^2(\theta) + r^2~sin^2(\theta) = r^2(cos^2(\theta) + sin^2(\theta))$.

-Dan