Chapter 10 - Practice Exercises - Page 594: 33

$r=-5 \sin \theta$

Here, we have $x^2+y^2+5y=0$ This implies that $x^2+(y+\dfrac{5}{2})^2=\dfrac{25}{4}$ Thus, we have an equation of a circle with center $(0,-\dfrac{5}{2})$ and radius $\dfrac{5}{2}$ So, $r^2+5r \sin \theta=0$ Hence, $r=-5 \sin \theta$