What is the polar form of the equation of (x+6)^2+y^2=36?

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What is the polar form of the equation of (x+6)^2+y^2=36?

r(r+12cosθ)=0
r^2=12rcosθ
r^2=12rsinθ
r(r+12sinθ)=0

Guest Apr 12, 2020

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2⁺⁰ Answers

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The answer is r^2 = 12*r*sin theta.

Guest Apr 12, 2020

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"What is the polar form of the equation of (x+6)^2+y^2=36?"

Alan Apr 12, 2020

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