## University Calculus: Early Transcendentals (3rd Edition)

Center $(0,\dfrac{3\sqrt 3}{2})$ and radius $\dfrac{3\sqrt 3}{2}$
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 $\dfrac{3\sqrt 3}{2}$