Question 1039812
<pre><b>First we get the rectangular equation of the circle.

The center is the polar point, which is

{{{(matrix(1,3,1,",",pi/2))}}}

That's the point that we get when we 
1. swing counter-clockwise {{{pi/2}}} 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)

{{{drawing(400,400,-4.5,4.5,-4.5,4.5, 
graph(400,400,-4.5,4.5,-4.5,4.5), 
green(circle(0,0,1),circle(0,0,2),circle(0,0,3),circle(0,0,4),circle(0,0,5),circle(0,0,6),circle(0,0,7)),
circle(0,1,0.15),circle(0,1,0.13),circle(0,1,0.11),circle(0,1,0.09),circle(0,1,0.07),circle(0,1,0.05),circle(0,1,0.03),circle(0,1,0.01) )}}}

So the circle with that center and a radius of {{{sqrt(5)}}} is this
black circle:

{{{drawing(400,400,-4.5,4.5,-4.5,4.5, circle(0,1,sqrt(5)),
circle(0,1,sqrt(5)-.02),circle(0,1,sqrt(5)+.02),
graph(400,400,-4.5,4.5,-4.5,4.5), circle(0,1,sqrt(5)),
green(circle(0,0,1),circle(0,0,2),circle(0,0,3),circle(0,0,4),circle(0,0,5),circle(0,0,6),circle(0,0,7)),
circle(0,1,0.15),circle(0,1,0.13),circle(0,1,0.11),circle(0,1,0.09),circle(0,1,0.07),circle(0,1,0.05),circle(0,1,0.03),circle(0,1,0.01) )}}}

which has rectangular equation:

{{{(x-h)^2+(y-k)^2=r^2}}} which becomes

{{{(x-0)^2+(y-1)^2=(sqrt(5))^2}}} or

{{{x^2+(y-1)^2=5}}}
{{{x^2+(y^2-2y+1)=5}}}
{{{x^2+y^2-2y+1=5}}}
{{{x^2+y^2-2y=4}}}

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

{{{drawing(200,100,-.5,2,-.5,1.5,line(-3,0,3,0),line(0,-3,0,3), line(0,0,1.5,1),line(1.5,0,1.5,1),locate(.4,.3,theta), locate(1.53,.62,y),locate(.77,0.05,x),locate(.75,.85,r) )}}} 

From which we see that 
{{{x/r=cos(theta)}}}, {{{y/r=sin(theta)}}}, and {{{x^2+y^2=r^2}}}

So we replace the x<sup>2</sup>+y<sup>2</sup> by r<sup>2</sup> and
the y using {{{y/r=sin(theta)}}} solved for y, {{{y=r*sin(theta)}}}

So the rectangular equation {{{x^2+y^2-2y=4}}} becomes
the polar equation:

{{{r^2-2r*sin(theta)=4}}}

Edwin</pre></b>