Question

Polar form of (x+6)^2 +y^2=36

Answer 1

r = -12cos(θ)

Explanation:

The usual translation can be used:

• x = r·cos(θ)
• y = r·sin(θ)

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