How do we find the closest point on the surface of the sphere to the given point outside of the sphere?
Well, my answer would be to construct a line between the centre and the given point and find the closer intersection. You can also do it using a unit vector of the given point and multiplying it by the sphere's radius. This makes it both intuitive and easy to show algebraically.
The confusing part for me is finding a point on a circle (formed by the intersection of a plane and a sphere) that is closest to the given point.
For example, consider a circle with a centre and radius: $$C = (6, 3, 6); r = 12 $$ respectively and a point outside $$P = (33, 30, 6)$$
I am only familiar with circles when we have $x$ and $y$ variables. Visualising this didn't help my conceptual understanding as I couldn't seem to find a way to express it.
Does anyone have any advice or tips on how to go about this? Thank you! I am adding an image to show what I mean and where the circle I mentioned is formed.
E.g.
If your intersecting plane (the one that determines the circle) has equation $$ \pi: Ax+By+Cz+D=0, $$ then its normal vector $N=(A,B,C)$ is parallel to the plane you need. If the point outside is $P(x_0,y_0,z_0)$ and the center of the sphere is $C(a,b,c)$ then the plane I mentioned in my comment has normal vector $$ N_1=N\times PC $$ and contains the point $C$ (or $P$). So its equation is $$ \pi_1: N_1\times (x-a,y-b,z-c)=0. $$ The points you are looking for solve the system formed by the two planes and the sphere. You get two points, the farthest and the closest.
By the way, the "equation" of the circle is a system of two equations, that of the sphere and that of $\pi$. From those, you can derive parametric equations as well.