Answer
Rectangular: $x^2+(y-4)^2=16$
Polar: $r=8\sin\theta$
Work Step by Step
The standard rectangular equation of a circle is:
$(x-h)^2+(y-k)^2=R^2$
Using the center's coordinates, we determine $h,k$:
$(h,k)=(0,4)$
$h=0$
$k=4$
We determine $r$ using the fact that the point $(0,0)$ belongs to the circle:
$(0-0)^2+(0-4)^2=R^2$
$R^2=16$
$R=4$
Determine the rectangular equation of the circle:
$x^2+(y-4)^2=16$
Determine the polar equation of the circle:
$x=r\cos\theta$
$y=r\sin\theta$
$(r\cos\theta)^2+(r\sin\theta-4)^2=r^2$
$r^2\cos^2\theta+r^2\sin^2\theta-8r\sin\theta+16=16$
$r^2(cos^2\theta+\sin^2\theta)-8r\sin\theta=0$
$r^2=8r\sin\theta$
$r=8\sin\theta$
