Write the rectangular equation (x-4)^2 + y^2=16 in polar form

Answer 1

x^2 - 8 x + 16 + y^2 = 16
or
x^2 - 8 x + y^2 = 0
let theta = T
x = r cos T
y= r sin T
r^2 cos^2 T - 8 r cos T + r^2 sin^2 T = 0
but cos^2+ sin^2 = 1
so
r^2 - 8 r cos T = 0
r = 8 cos theta

Answer 2

(x-4)^2 + y^2=16
x^2-8x+16+y^2=16
x^2-8x+y^2=0
since x^2+y^2=r^2,
r^2=8x
r^2=8rcos(theta)
r=8cos(theta)

I hope this helped

Answer 3

The above notes are true, but you know that
(x-4)^2 + y^2=16
is a circle of radius 4 with center at (4,0)

You also know that the circle r = 2a cosθ is a circle of radius a with center at (a,0) in x-y coordinates.

So, this one will be r = 8cosθ