Answer
A circle with center $(-3,0)$ and radius $3$
Work Step by Step
Consider $r=-6\cos \theta$
or, $r^2=-6r \cos \theta$
This implies that
$x^2+y^2=-6x$
This can be re-arranged as: $(x+3)^2+y^2=9$
Hence, we get the equation of a circle with center $(-3,0)$ and radius $3$