Answer
$(x-3)^2+y^2=9$
Work Step by Step
We are given the equation in polar coordinates:
$r=6\cos\theta$
Multiply both sides by $r$:
$r^2=6r\cos\theta$
Use the formulas:
$r^2=x^2+y^2$
$x=r\cos\theta$
We have:
$x^2+y^2=6x$
$x^2+y^2-6x=0$
$(x^2-6x+9)+y^2=0+9$
$(x-3)^2+y^2=9$
This is the equation of a circle with center $(3,0)$ and radius 3.
