Thread: 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.

Seriously, that's it.

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)

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?

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:







Now simplify. Note that
.

-Dan