Answer:
r = -12cos(θ)
Step-by-step explanation:
The usual translation can be used:
Putting these relationships into the formula, we have ...
(r·cos(θ) +6)² +(r·sin(θ))² = 36
r²·cos(θ)² +12r·cos(θ) +36 +r²·sin(θ)² = 36
r² +12r·cos(θ) = 0 . . . . subtract 36, use the trig identity cos²+sin²=1
r(r +12cos(θ)) = 0
This has two solutions for r:
r = 0 . . . . . . . . a point at the origin
r = -12cos(θ) . . . the circle of interest
