Is $r^2-4r\cos(\theta)=14$ an equation of a circle or cylinder?

Question

A question asks to identify the surface of the polar equation \begin{equation}\tag{1} r^2-4r\cos(\theta)=14. \end{equation} I converted $(1)$ into Cartesian coordinates: \begin{equation}\tag{2} (x-2)^2+y^2 = (3\sqrt{2})^2. \end{equation} I would have thought $(2)$ represents the equation of a cylinder, because the question asks to identify a surface.

Answer 1

This is of course a circle. General formula for the circle is $$(x-p)^2+(y-q)^2 =r^2$$

where $O=(p,q)$ is a center and $r$ radius. At your case $O=(2,0)$ and $r=3\sqrt{2}$

Answer 2

$$\begin{equation}\tag{1} r^2-4r\cos(\theta)=14. \end{equation}$$ $$x^2+y^2-4x=14$$ $$(x-2)^2+y^2=18$$ $$(x-2)^2+y^2=(3\sqrt{2})^2$$