# How do you convert x^ 2 + y ^2 + 3y = 0 into polar form?

Nov 12, 2016

$r = - 3 \sin \theta , \theta \in \left[- \pi , 0\right]$

#### Explanation:

The given equation ${x}^{2} + {y}^{2} + 3 y = 0$ can be reorganized as

${\left(x - 0\right)}^{2} + {\left(y + \frac{3}{2}\right)}^{2} = {\left(\frac{3}{2}\right)}^{2}$ that represents the circle with center at

cartesian $C \left(0 , - \frac{3}{2}\right) \mathmr{and}$ polar $C \left(\frac{3}{2} , - \frac{\pi}{2}\right)$

graph{x^2+y^2+3y=0 [-10, 10, -5, 5]}

Let O be the pole ( origin ) and P(r, theta) any point on the circle.

Then observing triangle OPC is isosceles and projecting the equal

sides OC and CP on OP,

$O P = r = 2 \left(r a \mathrm{di} u s\right) \cos \left(\theta - \left(- \frac{\pi}{2}\right)\right) = - 3 \sin \theta$.

Easy to find that $\theta \in \left[- \pi , 0\right]$ could make one full circle.