How do you convert #x^2+(y-4)^2=16# into polar form?

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Answer 1 · 2017-09-11T20:02:09

see below

Use the formulas #r^2=x^2+y^2, x= r cos theta, y= r sin theta#

#x^2+(y-4)^2 = 16# -->FOIL

#x^2+y^2-8y+16=16#

#x^2+y^2-8y=16-16#

#x^2+y^2-8y=0#

#(x^2+y^2)-8y=0#

#r^2-8rsin theta=0#