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

 

  • r^2=12rsinθ
  • r^2=12rcosθ
  • r(r+12cosθ)=0
  • r(r+12sinθ)=0
 Apr 22, 2020
 #1
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To change from rectangular to polar, replace  x with r·cos(theta)  and  y with r·sin(theta).

 

(x + 6)2 + y2  =  36     ----->     (  r·cos(theta)  +  6 )2  +  ( r·sin(theta) )2  =  36

Multiplying out:                r2·cos2(theta)  +  12·r·cos(theta)  +  36  +  r2·sin(theta)  =  36

Subtracting 36:                r2·cos2(theta)  +  12·r·cos(theta)  +   r2·sin(theta)  =  0

Rearranging:                    r2·cos2(theta)  +  r2·sin(theta)  +  12·r·cos(theta)  =  0

Factoring:                         r2( cos2(theta)  +  r2·sin(theta)  )  +  12·r·cos(theta)  =  0

Since cos2 + sin2  =  1:     r2 + 12·r·cos(theta)  =  0

Factoring:                          r( r + 12·cos(theta) )  =  0

 Apr 22, 2020

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