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

$r=-4 \cos \theta$
Here, we have $x^2+y^2+4x=0$ This implies that $(x+2)^2+y^2=4$ Thus, we have an equation of a circle with center $(-2,0)$ and radius $2$ So, $r^2+4r \cos \theta=0$ Hence, $r=-4 \cos \theta$