Thomas' Calculus 13th Edition

$r=-5 \sin \theta$
Consider $x^2+y^2+5y=0$ or, $x^2+(y+\dfrac{5}{2})^2=(\dfrac{25}{4})$ Here, we get the equation of a circle with center $(0,-\dfrac{5}{2})$ and radius $\dfrac{5}{2}$ Thus, $r^2+5r \sin \theta=0 \implies r=-5 \sin \theta$