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rcwha
06-05-2011, 07:53 PM
So there is a circle with a radius of 6 and a center of (0,0).

I understand that the equation for that is x[sup:2ujnllum]2[/sup:2ujnllum] + y[sup:2ujnllum]2[/sup:2ujnllum]= 36.

What i don't understand is the question which is......how does one return the x/y coordinates for any given radius and degree around the center point 0,0?

tkhunny
06-05-2011, 08:03 PM
Have you considered:

x = r\cos(\theta)

y = r\sin(\theta)

rcwha
06-05-2011, 08:23 PM
Have you considered:

x = r\cos(\theta)

y = r\sin(\theta)

Is this only valid for circles with a center of 0,0. With that being said, what exactly is the question asking?

tkhunny
06-05-2011, 10:35 PM
Other formulations are required for certer at some other location.

Check it out with r = 1 and \theta \in \left\{{0,\frac{\pi}{6},\frac{\pi}{4},\frac{\pi}{3 },\frac{\pi}{2}}\right\}

It should begin to make sense.

rcwha
06-06-2011, 12:00 AM
Other formulations are required for certer at some other location.

Check it out with r = 1 and \theta \in \left\{{0,\frac{\pi}{6},\frac{\pi}{4},\frac{\pi}{3 },\frac{\pi}{2}}\right\}

It should begin to make sense.

k i think i got it...so if radius is 6 and you want to know the coordinates at pi/6 , then you would multiply 6 *(sqrt3/(2)) for the x value(cos) and 6 x .5 for the y value(y).

Which goes with what you said in your last post. I pretty much need to understand/memorize the unit circle?

tkhunny
06-06-2011, 07:27 AM
Very good. And the unit circle is perfectly scalable. In this example, just multiply by 6.

rcwha
06-06-2011, 02:09 PM
Thanks for the help. Appreciate it!