+1993 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
+23245 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
