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

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Answer 1 · 2016-11-12T08:06:34

#r=-3sin theta, theta in [-pi, 0]#

The given equation #x^2+y^2+3y=0# can be reorganized as

#(x-0)^2+(y+3/2)^2=(3/2)^2# that represents the circle with center at

cartesian #C(0, -3/2) and# polar #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,

#OP = r = 2 (radius) cos (theta-(-pi/2))=-3 sin theta#.

Easy to find that #theta in [-pi, 0]# could make one full circle.