How do you convert #x^2+(y-3)^2=9# to polar form?

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Answer 1 · 2017-04-17T05:15:28

Please see the explanation.

Given: #x^2+(y-3)^2=9#

expand the square:

#x^2+y^2-6y+9=9#

Substitute #r^2" for "x^2+y^2#:

#r^2-6y=0#

Substitute #rsin(theta)# for y:

#r^2-6rsin(theta)=0#

We can discard a factor of r because the root #r = 0# is duplicated at #theta= 0#

#r-6sin(theta)=0#

#r = 6sin(theta)#