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+1
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avatar+1898 

What is the polar form of the equation?

(x+6)2+y2=36

 

r2=12rcosθ

 

r(r+12sinθ)=0

 

r2=12rsinθ

 

r(r+12cosθ)=0

 

is it D? im not sure, can someone explain

 Apr 23, 2020
 #1
avatar+20767 
+2

You are correct!

 

Here's how I looked at the problem:

 

x  =  r·cos(theta)          y  =  r·sin(theta)

 

(x + 6)2 + y2  =  36

( r·cos(theta)  + 6 )2 + ( r·sin(theta) )2  =  36

r2·cos2(theta) + 12·r·cos(theta) + 36  +  r2·sin2(theta)  =  36

r2·cos2(theta) +  r2·sin2(theta) + 12·r·cos(theta) + 36  =  36

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

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

r2·( 1 ) + 12·r·cos(theta)  =  0

r2 + 12·r·cos(theta)  =  0

r(1 + 12cos(theta)  =  0

 Apr 23, 2020

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