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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
 Apr 12, 2020
 #1
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The answer is r^2 = 12*r*sin theta.

 Apr 12, 2020
 #2
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"What is the polar form of the equation of (x+6)^2+y^2=36?"

 

 Apr 12, 2020

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