Kind of a bad thing to say about the UK isn't it?
Given point $\displaystyle C= (x_1, y_1, z_1)$ and [tex]P= (x_0, y_0, z_0)[/itex] then $\displaystyle <x- x_0, y_0, z_0>$ is a vector from the center of the sphere to C and so is normal to the sphere and a tangent plane at C. That means that we can write the equation of the plane as $\displaystyle (x_1- x_0)(x- x_0)+ (y_1-y_0)(y- y_0)+ (z_1-z_0(z- z_0)= 0$. Now, you say that M is a point on that tangent plane and it "coordinates" are known. But to give 3D coordinates, we would have to know what the coordinate system on that planeis and how it is connected to the three dimensional coordinate system.
I meant that I don't know where you would learn that in the US as the grading system is different.
To the problem, the coordinates are based between Cartesian coordinates which is what I need the product to be and this is all for a program. I think I have a solution by using the triangle PCM to get the angle MPC in both azimuth angle and inclination angle then getting a distance from the centre which I'm thinking about and then using these numbers and using the spherical coordinate system to get the Cartesian coordinates. I'll let you know if it works.