Given the coordinates of two sheaves $\overrightarrow{p_a}$ and $\overrightarrow{p_b}$ and the axis of rotation of their rotation $\overrightarrow{s_a}$(blue) and $\overrightarrow{s_b}$(cyan) respectively together with their radii |$\overrightarrow{r_a}$|(red) and |$\overrightarrow{r_b}$|(yellow) how do I find the lines tangent(green) to both sheaves, see the picture below. Which I believe should be four lines.
How can I obtain a definition for the tangent lines in the form of the 6 known ($\overrightarrow{p_a}$, $\overrightarrow{p_b}$, $\overrightarrow{s_a}$, $\overrightarrow{s_b}$, |$\overrightarrow{r_a}$| and |$\overrightarrow{r_b}$|).
Any help is more than welcome.
First, note that as depicted in the picture, the axes of rotation of the sheaves are parallel to $\left(\vec{s_a} - \vec{p_a}\right)$ and $\left(\vec{s_b} - \vec{p_b}\right)$ respectively, where $\vec{p_a}$ and $\vec{p_b}$ are the origins of the two sheaves. Similarly, we write the two radii as $\left\| \vec{r_a} - \vec{p_a} \right\|$ and $\left\| \vec{r_b} - \vec{p_b} \right\|$ respectively, where $\|\cdot\|$ denotes the norm.
Now, let $\vec{r_a}$ and $\vec{r_b}$ be the vectors corresponding to the radii of the two sheaves, such that they can be joined by a common tangent. So, $\vec{r_a}$ and $\vec{r_b}$ satisfy \begin{align} \left(\vec{r_a} - \vec{p_a}\right) \cdot \left(\vec{s_a} - \vec{p_a}\right) = 0 = \left(\vec{r_b} - \vec{p_b}\right) \cdot \left(\vec{s_b} - \vec{p_b}\right) , \end{align} because the axes of rotation are perpendicular to the corresponding radii. Next, the equation of a tangent at $\vec{r_a}$ to the first sheave is \begin{align} \vec{r_a} + t \vec{u} \;\text{ where }\; \left(\vec{r_a} - \vec{p_a}\right) \cdot \vec{u} = 0 , \end{align} since the tangent must pass through the vector $\vec{r_a}$ and be perpendicular to the radius vector $\left(\vec{r_a} - \vec{p_a}\right)$. Here, $t$ is any real number (scalar). So, a tangent to both sheaves passing through $\vec{r_a}$ and $\vec{r_b}$ would satisfy \begin{align} \left(\vec{r_a} - \vec{p_a}\right) \cdot \vec{u} &= 0, \\ \left(\vec{r_b} - \vec{p_b}\right) \cdot \vec{u} &= 0, \\ \vec{r_a} + t \vec{u} &= \vec{r_b} \end{align} for some scalar $t$, which implies \begin{align} &\left(\vec{r_a} - \vec{p_a}\right) \cdot \left(\vec{r_b} - \vec{r_a}\right) = 0 = \left(\vec{r_b} - \vec{p_b}\right) \cdot \left(\vec{r_b} - \vec{r_a}\right) . \end{align}
Therefore, we need to solve for $\vec{r_a}$ and $\vec{r_b}$ given $\vec{p_a}$, $\vec{p_b}$, $\vec{s_a}$, $\vec{s_b}$ and the two radii $r_a$ and $r_b$, such that \begin{align} &\left(\vec{r_a} - \vec{p_a}\right) \cdot \left(\vec{s_a} - \vec{p_a}\right) = 0 = \left(\vec{r_b} - \vec{p_b}\right) \cdot \left(\vec{s_b} - \vec{p_b}\right) ,\\ & \left(\vec{r_a} - \vec{p_a}\right) \cdot \left(\vec{r_b} - \vec{r_a}\right) = 0 = \left(\vec{r_b} - \vec{p_b}\right) \cdot \left(\vec{r_b} - \vec{r_a}\right) ,\\ & \left\|\vec{r_a} - \vec{p_a}\right\| = r_a , \;\text{ and }\; \left\|\vec{r_b} - \vec{p_b}\right\| = r_b . \end{align} The number of such tangents will depend on the existence and the number of solution pairs $\left(\vec{r_a}, \vec{r_b}\right)$ of the above equations. Given a solution pair $\left(\vec{r_a}, \vec{r_b}\right)$, the equation of the corresponding tangent joining $\vec{r_a}$ and $\vec{r_b}$ is $\left(t \vec{r_a} + (1 - t) \vec{r_b} \right)$, where $t$ is a real number (scalar).
The above procedure is applicable for general vector spaces with real valued scalars. Below, it is reduced to the space $\mathbb{R}^3$ of three dimensional vectors.
Let us denote a vector $\vec{u}$ in $\mathbb{R}^3$ as $(u_1, u_2, u_3)$. So, in this notation, $\vec{r_a} = (r_{a,1}, r_{a,2}, r_{a,3})$, $\vec{r_b} = (r_{b,1}, r_{b,2}, r_{b,3})$, $\vec{p_a} = (p_{a,1}, p_{a,2}, p_{a,3})$, $\vec{p_b} = (p_{b,1}, p_{b,2}, p_{b,3})$, $\vec{s_a} = (s_{a,1}, s_{a,2}, s_{a,3})$, $\vec{s_b} = (s_{b,1}, s_{b,2}, s_{b,3})$. And the above set of equations reduces to \begin{align} & (r_{a,1} - p_{a,1}) (s_{a,1} - p_{a,1}) + (r_{a,2} - p_{a,2}) (s_{a,2} - p_{a,2}) + (r_{a,3} - p_{a,3}) (s_{a,3} - p_{a,3}) = 0 ,\\ & (r_{b,1} - p_{b,1}) (s_{b,1} - p_{b,1}) + (r_{b,2} - p_{b,2}) (s_{b,2} - p_{b,2}) + (r_{b,3} - p_{b,3}) (s_{b,3} - p_{b,3}) = 0 ,\\ & (r_{a,1} - p_{a,1}) (r_{b,1} - r_{a,1}) + (r_{a,2} - p_{a,2}) (r_{b,2} - r_{a,2}) + (r_{a,3} - p_{a,3}) (r_{b,3} - r_{a,3}) = 0 ,\\ & (r_{b,1} - p_{b,1}) (r_{b,1} - r_{a,1}) + (r_{b,2} - p_{b,2}) (r_{b,2} - r_{a,2}) + (r_{b,3} - p_{b,3}) (r_{b,3} - r_{a,3}) = 0 ,\\ & (r_{a,1} - p_{a,1})^2 + (r_{a,2} - p_{a,2})^2 + (r_{a,3} - p_{a,3})^2 = r_a^2 ,\\ & (r_{b,1} - p_{b,1})^2 + (r_{b,2} - p_{b,2})^2 + (r_{b,3} - p_{b,3})^2 = r_b^2 . \end{align} There are six equations and six unknowns to solve: $r_{a,1}, r_{a,2}, r_{a,3}, r_{b,1}, r_{b,2}, r_{b,3}$. Since four of these equations are quadratic equations, there will be multiple sets of solutions if solutions exist.