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

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Answer 1 · 2016-11-12T08:06:34

$r = - 3 sin θ, θ \in [- π, 0]$ $r=-3sin theta, theta in [-pi, 0]$

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

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

cartesian $C (0, - \frac{3}{2}) and$ $C(0, -3/2) and$ polar $C (\frac{3}{2}, - \frac{π}{2})$ $C(3/2, -pi/2)$

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,

$O P = r = 2 (r a d i u s) cos (θ - (- \frac{π}{2})) = - 3 sin θ$ $OP = r = 2 (radius) cos (theta-(-pi/2))=-3 sin theta$ .

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

How do you convert x^ 2 + y ^2 + 3y = 0x2+y2+3y=0x^ 2 + y ^2 + 3y = 0 into polar form?