Answer
A circle with center $(0,\dfrac{3\sqrt 3}{2})$ and radius $\dfrac{3\sqrt 3}{2}$
Work Step by Step
Consider $r=3\sqrt 3\sin \theta$
or, $r^2=(3\sqrt 3) r \sin \theta$
This implies that $x^2+y^2-3\sqrt 3y=0$
This can be re-arranged as:
$x^2+(y-\dfrac{3\sqrt 3}{2})^2=\dfrac{27}{4}$
Hence, we get the equation of a circle with center $(0,\dfrac{3\sqrt 3}{2})$ and radius $\dfrac{3\sqrt 3}{2}$
