SOLUTION: Determine a polar equation for the circle with a radius of sqrt(5) and a center at (1,pi/2).

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Question 1039812: Determine a polar equation for the circle with a radius
of sqrt(5) and a center at (1,pi/2).

Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
First we get the rectangular equation of the circle.

The center is the polar point, which is

%28matrix%281%2C3%2C1%2C%22%2C%22%2Cpi%2F2%29%29

That's the point that we get when we 
1. swing counter-clockwise pi%2F2 radians or 90° 
   from the right side of the x-axis, and
2. Go out 1 unit in that direction, which
   turns out to be directly upward.

So that's the same point as the rectangular point (0,1)



So the circle with that center and a radius of sqrt%285%29 is this
black circle:



which has rectangular equation:

%28x-h%29%5E2%2B%28y-k%29%5E2=r%5E2 which becomes

%28x-0%29%5E2%2B%28y-1%29%5E2=%28sqrt%285%29%29%5E2 or

x%5E2%2B%28y-1%29%5E2=5
x%5E2%2B%28y%5E2-2y%2B1%29=5
x%5E2%2By%5E2-2y%2B1=5
x%5E2%2By%5E2-2y=4

Then to change from rectangular to polar or
vice-versa we draw this:

 

From which we see that 
x%2Fr=cos%28theta%29, y%2Fr=sin%28theta%29, and x%5E2%2By%5E2=r%5E2

So we replace the x2+y2 by r2 and
the y using y%2Fr=sin%28theta%29 solved for y, y=r%2Asin%28theta%29

So the rectangular equation x%5E2%2By%5E2-2y=4 becomes
the polar equation:

r%5E2-2r%2Asin%28theta%29=4

Edwin