Answer
$ r=16\cos\theta$
Work Step by Step
The circle passes through the origin, as (0,0) satisfies the Cartesian equation.
Completing the square,
$(x^{2}-16x+8^{2})+y^{2}=0+8^{2}$
$(x-8)^{2}+y^{2}=64$
The radius is $8$, and the center is at $(8,0)$.
In polar coordinates, the center lies at $r_{0}=8, \theta_{0}=0,\qquad P(8,0)$
A circle passing through the origin, of radius $a$, centered at $P_{0}(r_{0}, \theta_{0}),$
has the polar equation
$r=2a\cos(\theta-\theta_{0})$
So this circle has equation
$r=2(8)\cos(\theta-0)$
or
$ r=16\cos\theta$
