Being given a general quadric with implicit equation:
$$\tag{1}ax^2+by^2+cz^2+2dxy+2eyz+2fzx+2gx+2hy+2iz+k=0,$$
the key point is that the tangent plane to this quadric in $(x',y',z')$ is obtained by "polarization", i.e., by looking at the homogeneous version of (1). This homogeneous form is obtained by introducing a 4th variable $t$, atop variables $x,y,z$, in such a way that if a term is only 1st degree, it is multiplied by $t$, and if it is a constant, it is multiplied by $t^2$: thus, all terms become of degree 2 (whence the name "homogenization"), in this way:
$$\tag{1}ax^2+by^2+cz^2+2dxy+2eyz+2fzx+2gtx+2hty+2itz+kt^2=0$$
Remark: if $t=1$, one finds back (1).
Then the method amounts to build the bilinear form associated with this quadratic form in 4 variables:
$\tag{2}axx'+byy'+czz'+d(x'y+xy')+e(y'z+yz')+f(z'x+zx')+g(t'x+x't)+h(t'y+ty')+i(t'z+tz')+ktt'=0.$
Remark: if all the "primes" are suppressed, we are back to equation (2).
Then make $t=t'=1$ in (2),
$$\tag{3}axx'+byy'+czz'+d(x'y+xy')+e(y'z+yz')+f(z'x+zx')+g(x+x')+h(y+y')+i(z+z')+k=0,$$
We have obtained in (3) the equation of the tangent plane to the quadric at point $(x',y',t')$.
Remark: this techniques can be justified in the framework of projective geometry.
Applying this to our case, we get, for the tangent plane at $(x_0,y_0,z_0)$ is
$$xx_0+4yy_0+zz_0=9$$
This is the equation of a plane with normal vector
$$\pmatrix{x_0\\4y_0\\z_0} \text{which is desired to be proportional to} \pmatrix{-4\\8\\-2}.$$
Thus, we must look for a coefficient $\alpha$ such that:
$$\tag{4}\pmatrix{x_0\\4y_0\\z_0}=\pmatrix{-4\alpha\\8\alpha\\-2\alpha}$$
Point $(x_0,y_0,z_0)$ is a point of the ellipsoid if and only if:
$$(-4\alpha)^2+4(2\alpha)^2+(-2\alpha)^2=9$$
i.e., $$\alpha^2=\frac{9}{36}=\frac{1}{4} \ \ \Leftrightarrow \ \ \alpha=\pm \frac{1}{2}$$
Whence the two solution points by plugging these opposite values of $\alpha$ in (4):
$$\pmatrix{x_0\\y_0\\z_0}=\pm \frac{1}{2} \pmatrix{-4\\2\\-2}=\pm \pmatrix{-2\\1\\-1}.$$
As you say, the tangent vector to a circle is always perpendicular to the radial vector. Since you know the radial vectors at the two points, finding the corresponding direction vectors is a simple matter of rotating the radial vectors 90° in the appropriate direction—the same direction in which you’re measuring the angle of arc.
Assuming the standard mathematical convention of angles measured counterclockwise begin positive, if the radial vector from $C$ to a point has coordinates $(a,b)$, then the direction vector at that point is $(-b,a)$ for a counterclockwise motion along the arc and $(b,-a)$ for a clockwise motion. Adjust the length of this vector as desired.
If you don’t happen to have the direction of motion handy, you can determine it from a pair of points near each other along the path. Let $(x_1,y_1)$ and $(x_2,y_2)$ be the coordinates of the radial vectors from $C$ to these two points, with the latter point being reached later in time. Compute the cross product $$(x_1,y_1,0)\times(x_2,y_2,0) = (0,0,x_1y_2-x_2y_1).$$ By the right-hand rule, the sign of the last coordinate of the result will give you the direction of motion: positive for counterclockwise, negative for clockwise. You want the two points to be near each other so that the direction is computed correctly. If they’re too far apart, the direction computed by the cross product will be the reverse of the actual direction of motion, as would happen if you used the two end points in your illustration.
Best Answer
$(a,b)$ is a point but if you view $\mathbb R^{2}$ as a vector space, the square of the magnitude of $(a,b)$ is $ |(a,b)|^{2} = a^2 + b^2$. That's why $x^2+y^2 = a^2+b^2$ defines a circle.
If $F = \sqrt{a^2 + b^2}$, this just says that that the magnitude of the vector field at $(a,b)$ is equal to the distance from $(a,b)$ to $(0,0)$.
The tangent to a circle is perpendicular to the radius of the circle. If $(x,y)$ is any point in the plane, we can represent it as $(x,y) = x(1,0) + y(0,1) = xi + yj$. A vector $v$ perpendicular to this must be such that $v\cdot (x,y) = 0$. We also require the magnitude to equal $\sqrt{x^2 + y^2}$. There are two obvious candidates, $(-y,x)$ and $(y,-x)$. One of these points in the clockwise direction and the other points in the counter-clockwise direction. If you draw a point in the first quadrant along with the two candidate vectors you'll see that $yi + -xj$ is the clockwise one.