If the velocity vector of a moving particle is always perpendicular to the position vector, is the path a circle? A Newtonian physics question:
If the velocity vector of a moving particle is always perpendicular to the position vector, is the only possible path a circle? 
What if the magnitude of the velocity vector changes, what if it doesn't?
EDIT: I meant to write "perpendicular" not "tangent"
ANOTHER EDIT: UGH, I'm sorry, I meant to say this:  If the velocity vector of a moving particle is always perpendicular to the position vector, is the only possible path a circle?  What if the magnitude of the velocity vector changes?  
 A: "Velocity always perpendicular to position vector" means that the distance from the particle of interest to the origin never changes. The question is whether such motion always traces out a circle.
Counterexample: consider a plane pendulum with its pivot at the origin. The velocity always obeys $\vec v \cdot \vec r = 0$, which is a good definition of "perpendicular," but the motion only traces out a small arc of a circle.
Better counterexample, if you don't allow the velocity to pass through zero: allow your pendulum to move in three dimensions. The particle will always lie on the surface of a sphere, but you can trace out arbitrary paths on the sphere — you're not limited to great circles. For instance, if I brought you to my house and gave you a sparkler, you'd probably hold your arm out stiff (distance from shoulder to sparkler = constant) and spell out your name. (I'm assuming your name isn't Oooo.)
A: No its not always a circle. For example, in 3 dimension, a particle can be constrained to move on a spherical surface. It can describe any path on the surface but the velocity vector is always tangential to the surface or perpendicular to the position vector from the center of the sphere.
A: Mathematically, yes, it will always be a circle. If you look in polar coordinates, your velocity vector is $\vec{v}=v(t)\hat{\theta}$. It does not really matter what this velocity is, because no velocity in the radial direction, means no movement in that direction.
The question whether this is still physical? Probably not if the magnitude is changing. You have to take into account conservation of angular momentum for example, where external forces play a big role if $\vec{v}$ is changing.
A: 
I meant to ask the question: " If the velocity vector of a moving particle is always perpendicular to the position vector, is the only possible path a circle? What if the magnitude of the velocity vector changes?"

If $\vec r \times \vec v$ always points in the same direction, the particle is restricted to moving along a circular arc. It's not necessarily a circle, as rob's answer shows.
That isn't the only choice of "velocity vector of a moving particle is always perpendicular to the position vector". Space is three dimensional. What if the direction of the velocity vector changes, but always remains orthogonal to the position vector? The particle is now restricted to motion along a spherical surface.
I wrote the above in the context of Newtonian mechanics. Things get just a bit more interesting in relativistic mechanics where the dot product isn't Euclidean.
A: In 3-dimensional space, the particle is restricted to moving inside a spherical shell centred about the origin of the position vector, i.e. that particle will remain at a fixed distance from the origin of the position vector. In 2-D space, the motion of the particle s restricted to that of a circle. Here's my proof:
Let's say we have a position vector $\vec r$. Therefore, the velocity is $\frac{d\vec r}{dt}$.
The two vectors, position and velocity, are perpendicular, so the scalar product of the two vectors is zero. This is due to the geometric definition of the scalar product: $\vec a\cdot \vec b = \left|\vec a\right|\left|\vec b\right|\cos\theta$, where $\theta$ is the angle between the two vectors.
$$\vec r \cdot \frac{d\vec r}{dt} = 0$$
By using a nice trick, this is actually equivalent to:
$$\frac{d}{dt}\left(\vec r \cdot \vec r\right) = 0$$
This becomes apparent whenever you use the product rule for scalar products on the new expression:
$$\frac{d}{dt}\left( \vec r \cdot \vec r \right) = \vec r \cdot \frac{d\vec r}{dt} + \frac{d\vec r}{dt} \cdot \vec r = 2\vec r \cdot \frac{d\vec r}{dt}$$
$$2\vec r \cdot \frac{d\vec r}{dt} = 0 \rightarrow \vec r \cdot \frac{d\vec r}{dt} = 0$$
Now, from our second equation, we can integrate both sides wrt time to get:
$$\vec r \cdot \vec r = constant$$
By looking at the geometric definition of the scalar product, and noting that the two $\vec r$ vectors must of course be parallel, then it turns out that:
$${\left| \vec r \right|}^2 = constant$$
$$\therefore \left| \vec r \right| = constant$$
That is, the distance of the particle from the origin is fixed!
