Is the vector field normal or tangential to the curve? Given the curve $C$, $C = {(x,y):x^2+y^2=1}$, $n=\langle x,y\rangle$ is normal to $C$. 
Consider the vector field $F$ defined by $F=\langle y,-x\rangle$.
Is the vector field $F$ tangent to $C$ or normal to $C$ at points on $C$?
 A: It might be helpful to visualize what's happening here: for example, graphing the unit circle ($x^2 + y^2 = 1$) and the vector field $F$. Then consider what is happening at given points of the circle, e.g. at $(1, 0)$, and $(0, 1$. 
In your comment above you observed that the vectors in $F$ go in "circle style". I think what you are observing each vector in $F$ is tangent to $C$, and tangent at some point $(x, y)$ of $C$, with each vector directed counter-clockwise.

We know that for each point $(x, y)$ that lies on $C$, the vector $n=\langle x, y\rangle$ is normal to $C$ (it's a given) at that point, and so at the point $(1, 0)$, $n$ lies along the $x$-axis, pointing in the positive $x$ direction. The vectors in $F:\;$ are each orthogonal to the corresponding vectors $n$, as you point out in your comment above. So the vector $f\in F$ is perpendicular to the x-axis at that point, and hence, must be tangent to $C$ at $(1, 0)$. 
Likewise, consider what is happening at $(0, 1)$: At this point, $n$ lies along the y-axis, directed in positive $y$ direction (and normal to $C$). $f \in F$, at this point on $C$ is a "horizontal" vector, directed counter-clockwise, and hence tangent to $C$ at $(0, 1)$  
You are correct that the vectors in $F = \langle -y, x \rangle$ are orthogonal to the corresponding vectors $n = \langle x, y \rangle$. We just need to know what's happening with vectors in $F$ at points of $C$, knowing $n$ is normal to $C$.
Included immediately below is a representation of vectors $n = \langle x, y\rangle$ for $(x,y) \in C$, normal to $C$:
$\quad\quad\quad\quad\quad$ 
See the image below for an representation of vectors tangent to the unit circle, representing vectors in $F$:
$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$
A: You should memorize once and for all that, if ${\bf r}:=(x,y)$ is a vector in ${\mathbb R}^2$ attached to $(0,0)$, then ${\bf r}':=(-y,x)$ is the vector obtained by rotating ${\bf r}$ ninety degrees counterclockwise, in other words: ${\pi\over2}$ in the positive direction.
Given the circle $C:\ x^2+y^2=1$ we all know that at each point ${\bf r}:=(x,y)\in C$ the vector ${\bf r}$, attached to $(x,y)$, points in the outward radial direction vs. $C$, which is orthogonal to the tangential direction of $C$ at $(x,y)$. It follows that the vector ${\bf r}':=(-y,x)$ attached to $(x,y)$ points in the tangential direction of $C$ at $(x,y)$, to be exact: in the "positive" direction of $C$ when $C$ is described counterclockwise. Therefore your vector ${\bf F}=-{\bf r}'$ is tangential to $C$ as well at $(x,y)$.
