Mapping Coordinates on a Plane Tangent to a Sphere into Cartesian Coordinates in 3D Space

Question

Before we begin, I must ask you to keep the vocabulary at high school level.

These variables define the point the plane needs to be tangent to – the center of the circle is at the origin.
r - defines the radius of the sphere
Ø₁ – defines the rotation in the x-z axis – can be from 0 to 2π
Ø₂ – defines the rotation in the y axis – can be from 0 to π

I have found that I can convert that point on the sphere to Cartesian coordinates using equations.
x₀=r•cosØ₁•sinØ₂
z₀=r•sinØ₁•sinØ₂
y₀=r•cosØ₂

These variables define the location of the point to be converted on the 2d plane that is tangent to the sphere.
d₁ – is the x coordinate
d₂ – is the y coordinate

Positive y for the 2d plane is in the vertical direction as much as possible (because the plane is tangent to the sphere), positive x is in the counterclockwise direction looking down on the sphere from positive y in 3d space (or to the right if the plane is on your side of the sphere).

The formulas I have made are:

x₁=x₀+d₂•cosØ₁•cosØ₂-d₁•sinØ₁
z₁=z₀+d₂•sinØ₁•cosØ₂+d₁•cosØ₁
y₁=y₀+d₂•sinØ₂

At this point I hope you understand what I am trying to do and can point me to where I’ve gone wrong.

Answer 1

The first thing I did is I changed the axis to a more conventional format.

All I really did is switch the y and z axis.

To that effect, this is the revised polar-to-cartesian coordinate conversion equations.
x₀=r•cosØ₁•sinØ₂
y₀=r•sinØ₁•sinØ₂
z₀=r•cosØ₂

This is the plane that is tangent to the sphere.

These are the equations that convert coordinates on the plane into 3d cartesian coordinates.
And the written version (in case you can't load images for some reason)
x₁=x₀-d₂•cosØ₁•cosØ₂-d₁•sinØ₁
y₁=y₀-d₂•sinØ₁•cosØ₂+d₁•cosØ₁
z₁=z₀+d₂•sinØ₂

This is a close-up on the diagrams.

Answer 2

So you are asking to find the equation of the plane (or points $(x_1,y_1,z_1)$ on the plane) that is tangent to the sphere at the point $(x_0,y_0,z_0)$.

I will following the regular right hand rule, so the $z$ axis in your first picture should point outward. The $\theta_2$ starts from the positive $y$ direction.

So your $(x_0,y_0,z_0)$ in spherical coordinates is correct. Then you find the directive with respect to $theta_1,theta_2$ to find the tangent direction.

The new points are

$$ x_1=x_0+d_2\cos{\theta_2}\cos{\theta_1}-d_1\sin{\theta_2}\sin{\theta_1}\\ z_1=z_0+d_2\cos{\theta_2}\sin{\theta_1}+d_1\sin{\theta_2}\cos{\theta_1}\\ y_1=y_0-d_1\sin{\theta_2}$$