Answer
Center $(0,\dfrac{3\sqrt 3}{2})$ and radius $\dfrac{3\sqrt 3}{2}$
Work Step by Step
Here, we have $r=3\sqrt 3\sin \theta$
This implies that
$r^2=3\sqrt 3r \sin \theta$
$x^2+y^2-3\sqrt 3y=0$
or, $x^2+(y-\dfrac{3\sqrt 3}{2})^2=\dfrac{27}{4}$
Thus, we have a equation of a circle with center $(0,\dfrac{3\sqrt 3}{2})$ and radius $\dfrac{3\sqrt 3}{2}$
