How to write the polar equation for these rectangular equations?

Question

How would I find the polar equation for y=5? 
What about the polar equation for (x-6)^2 + y^2 = 36 ? 

Answer 1

The equation y = 5

Substitute for y = r sin(θ)

r sin(θ) = 5

r = 5/sin(θ)

Polar form of y = 5 is r = 5/sin(θ).

Answer 2

The circle equation (x - 6)² + y² = 36

x² + 36 - 12x + y² = 36

x² + y² - 12x = 36 - 36

x² + y² - 12x = 0

Substitute x = r cos(θ) and x² + y² = r² in above equation.

r² - 12r cos(θ) = 0

Factor out r.

r[r - 12 cos(θ)] = 0

r = 0 and r - 12 cos(θ) = 0

We omit the first part r = 0, because the equation r = 0 is the pole.

we therefore can keep only the second equation r - 12 cos(θ) = 0

r = 12 cos(θ)

The equation in polar form r = 12 cos(θ).