I was able to figure the substitutions inside the equation, but I'm stuck with the equation's manipulation that will give me the solution. What would be my next step?
$$r = 4\cos\theta$$
$$r^2 = 4r\cos\theta \tag{multiply both sides by $r$}$$
$$x^2+y^2 = 4x \tag{$r^2=x^2+y^2\quad r\cos\theta=x$}$$
$$x^2+y^2-4x=0 $$
Best Answer
Since you've already done most of the work and undoubtedly the hardest part of it, the secret is: complete the square:
$$x^2+y^2-4x=0\iff (x-2)^2+y^2=4$$