I read somewhere that the point of tangency of is that point for which the distance between it and the center is smaller than the distance between the center and any other point on the tangent plane. I tried proving this by using Lagrange Multipliers to minimize the distance between the center of the sphere $(x-a)^2 + (y-b)^2 + (z-c)^2=r^2$ and points on some tangent plane, but this was becoming extraordinarily tedious, so much so that I abandoned the approach. Perhaps there is a way of reducing it the case in which the center is the origin.
Here is something interesting, however. Letting $d(x,y,z) = (x-a)^2 + (y-b)^2 + (z-c)^2$ , where $(x,y,z)$ is a point on the tangent plane, if we assume that $(x,y,z)$ is the point of tangency, then $d(x,y,z) = r^2$. Now, if there were to exist some other point $(x,y,z)$ such that $g(x,y,z) < r^2$, then $(x-a)^2 + (y-b)^2 + (z-c)^2 < r^2$, which says that the point is inside the sphere. Since $g$ is a continuous function from the connected space $\Bbb{R}^3$ to $\Bbb{R}$ in the order topology, we should have by the generalized intermediate theorem that the tangent plane intersects the sphere at no fewer than two points, which cannot happen.
I am not sure if my reasoning is correct; I think I need to explicitly show there is another point $(p,q,r)$ on the tangent plane such that $(p-a)^2 + (q-b)^2 + (r-c)^2 > r^2$. Hopefully someone can point out the mistake or complete the ideas.
I'd take other approach.
Let $C$ be the center of the sphere and $T$ the point of tangency. Let $X$ be any other point of the plane.
Then the triangle $CTX$ is right and $CX$ is its hypotenuse. This implies $CX>CT$.