Answer
$x^2+(y-\frac{1}{6})^2=\frac{1}{36}$
circle with center $(0,\frac{1}{6})$, radius $r=\frac{1}{6}$
See graph.
Work Step by Step
1. $3r=sin\theta\Longrightarrow 3r^2=r\ sin\theta \Longrightarrow 3(x^2+y^2)=y \Longrightarrow x^2+y^2-\frac{1}{3}y=0 \Longrightarrow x^2+(y-\frac{1}{6})^2=\frac{1}{36}$
2. We can identify the equation as a circle with center $(0,\frac{1}{6})$, radius $r=\frac{1}{6}$
3. See graph.
