I assume by the curve's 'trace' you mean its image in Euclidean space.
(1) Let the fixed point be $\mathbf{p}$, and parametrize the curve by $\mathbf{x}(t)$. If the displacement vector going from $\mathbf{x}$ to $\mathbf{p}$ is parallel to the normal of the curve, then it must also be orthogonal to the tangent vector, and the tangent vector is parallel to $\mathbf{x}'(t)$. So we have
$$(\mathbf{x}(t)-\mathbf{p})\cdot\mathbf{x}'(t)=0 $$
The left hand side is actually the derivative of a simple expression. Thus, integrating yields
$$\frac{1}{2}\|\mathbf{x}(t)-\mathbf{p} \|^2=C.$$
This is precisely the equation of a circle of radius $\sqrt{2C}$ centered at $\mathbf{p}$.
(2) The curve here will actually be an open ray emanating outward from $\mathbf{p}$, if I'm understanding the question correctly. If the displacement $\mathbf{x}(t)-\mathbf{p}$ is parallel to the tangent vector, and so is $\mathbf{x}'(t)$, and the curve is regular, then there is a scalar function $\alpha(t)$ such that
$$\alpha(t)(\mathbf{x}(t)-\mathbf{p})=\mathbf{x}'(t).$$
This can be solved with the integrating factor method, but we'll take a shortcut. Reparametrize the curve so that the velocity is exactly $\mathbf{x}(s)-\mathbf{p}$, and then define $\mathbf{y}(s)=\mathbf{x}(s)-\mathbf{p}$. Then
$$\frac{d\mathbf{y}}{ds}=\mathbf{y}(s)$$
whose only solution is $\mathbf{y}(s)=e^s\mathbf{a}$ for some constant vector $\mathbf{a}$, which means that $\mathbf{x}(s)=\mathbf{p}+e^s\mathbf{a}$.
Fix $u \equiv u_0$. At a point $(u_0, v)$, the tangent plane to the surface at $x(u_0, v)$ is spanned by the vectors $\alpha_1'(u_0)$ and $\alpha_2'(v)$. This means that any line with a direction vector $\alpha_1'(u_0)$ will be parallel to all tangent planes along $u \equiv u_0$.
Best Answer
Sounds like you have a good strategy. You are looking at the cross section in some plane $z=z_0$, and you want to prove that the cross section here is a circle.
You are given that any normal line will intersect the $z$ axis. Let's write this line as $\vec{p}+\vec{n}t$ where $\vec{n}$ is the normal vector at the point $\vec{p}$ we are discussing. You can then show that a new line $\vec{p} + (\vec{n}-\hbox{proj}_{\hat{z}}\vec{n})t$ will also intersect the $z$ axis (taking away the z component doesn't change the fact that the line is in a radial direction).
Moreover, this new line will be normal to the cross section at $\vec{p}$: the tangent plane projects to a tangent line of the cross section, so must a normal vector of the surface project to a normal vector of the cross section.
This gives you the criteria you wanted -- all horizontal lines orthogonal to cross sections will intersect at the origin. This tells you roughly that the cross sections are circular, and that the overall shape will be a surface of revolution.