This is a confusion that I've always had, that I've never quite gotten rid of. (I'm definitely making a foolish mistake somewhere, I know... Sorry) Suppose $x,v \in \mathbb{R}^3$, and $c \in \mathbb{R}$. The equation of a plane in $\mathbb{R}^3$ is given: $\{x: \langle x, v\rangle=c\}$, where the inner product denotes the dot product. The reason why this is a plane is, for an $x_0$ such that $\langle x_0, v\rangle=c$, we have $\langle x-x_0, v\rangle=0$, so $x-x_0$ is a plane that is perpendicular to the vector $v$.
But alternately, using $x \cdot v = |x| |v| \cos (\theta)$ where $\theta$ is the angle between $x,v$, shouldn't $\{x: \langle x, v\rangle=c\}$ look like a circle around the vector $v$ in mid air (i.e. away from the origin of $\mathbb{R}^3$), where the vector $x_0$ (defined as above) has been rotated around the vector $v$? Because rotating the vector $x_0$ around the vector $v$, the angle between $x_0$ and $v$ stays equal, while there is no change in the length of the two vectors. But clearly this is wrong, as it must be a plane, not a circle.
In the situation you ask about where $\,P := \{x: \langle x, v\rangle=c\}\,$ is the equation of a plane in $\,\mathbb{R}^3,\,$ there exists a unique point $\,O\in P\,$ where the angle $\,\theta\,$ between $\,O\,$ and $\,v\,$ is $\,0.\,$ This point $\,O\,$ is the origin in a polar coordinate system for the plane $\,P.\,$ The points $\,x\,$ with a constant angle between them and $\,v\,$ are a constant distance from $\,O\,$ in the plane and hence they form a circle. Conversely, all circles in the plane $\,P\,$ with a fixed distance from the origin $\,O\,$ are described by $\,\{x\in P: |x| = d\}\,$ for some $\,d\,$ or alternatively $\,\{x\in P: |x-O| = r\}\,$ for some $\,r.\,$