I have a similar problem to this problem. I have to external points A and B and a sphere with radius r. I want to find out the point of tangency T. Below is the illustration:
So far I have applied the sine rule to find out the angle CAT, but not sure on how to make use of it.
Any help is appreciated.
On
The “point” of tangency to a sphere is a circle, specifically, the intersection of the polar plane to $A$ and the sphere. If you have a specific tangent line in mind, then its point of tangency is the intersection of the line and this plane.
Working in homogeneous coordinates and denoting points with lower-case letters, we write the equation of the sphere in matrix form: $$\mathbf x^T C \mathbf x = \begin{bmatrix}x&y&z&1\end{bmatrix} \left[\begin{array}{c|c} I_3 & -\mathbf c \\ \hline -\mathbf c^T & \mathbf c^T\mathbf c-r^2 \end{array}\right] \begin{bmatrix}x\\y\\z\\1\end{bmatrix} = 0.$$ The polar plane of the point $\mathbf a$ is then $C\mathbf a$ and the intersection with the line through $\mathbf a$ and $\mathbf b$ can be computed directly using the Plücker matrix $\mathbf a\mathbf b^T-\mathbf b\mathbf a^T$ of the line as $$\mathbf t = (\mathbf b^TC\mathbf a)\mathbf a - (\mathbf a^TC\mathbf a)\mathbf b.$$ Divide through by the last coordinate to convert back into inhomogeneous Cartesian coordinates.
The fact that $T$ lies on the tangent line is characterised by the existence of a real number $t$ such that $$\left(\matrix{x_3\\y_3\\z_3}\right)=\left(\matrix{tx_1+(1-t)x_4\\ty_1+(1-t)y_4\\tz_1+(1-t)z_4}\right)$$
The fact that it is the tangency point is characterised by $CT$ being orthogonal to $AB$. If you write down $$\vec{CT}\cdot\vec {AB}=0$$ you get a linear equation of degree $1$ in $t$.
The solution is $$t=\frac{(x_2-x_4)(x_1-x_4)+(y_2-y_4)(y_1-y_4)+(z_2-z_4)(z_1-z_4)}{(x_1-x_4)^2+(y_1-y_4)^2+(z_1-z_4)^2}$$