Premise
I am trying to solve the following problem:
We have two spheres of given radius $a.$ Their centers $C_1$ and $C_2$ lie on a line $d$ given by its parametric equation, which can be compactly written as $\mathbf{r}=\mathbf{r_0}+t\mathbf{u}$ where $t\in\mathbb R$ and $\mathbf{u}$ denotes the direction vector of the line.
Furthermore, there is a plane $P,$ whose normal vector is denoted by $\mathbf n,$ that is tangent to both spheres.
- Determine the coordinates of the centers of the two spheres.
My approach
- Let's call the tangent points between the plane and the two spheres $P_1$ and $P_2$ respectively. Then we know the vector from $C_1$ to $P_1$ is orthogonal to the plane, and thus it is a multiple of the normal vector $\mathbf n.$
- Moreover, we know the radius which here should also coincide with the distance between $C_1$ to the plane. So we can write
$$\mathbf{C_1-P_1}=a\frac{\mathbf n}{||\mathbf{n}||} \tag{1}$$ Similarly, for $C_2:$ $$\mathbf{C_2-P_2}=a\frac{\mathbf n}{||\mathbf{n}||} \tag{2}$$
- Then we also have the information that the centers of the two spheres also lie on a line $d$, which gives us yet another condition, since the vector of $\mathbf{C}_2-\mathbf{C}_1$ should in fact lie on the line, thus it is proportional to the direction vector $\mathbf u:$
$$ \mathbf{C}_2-\mathbf{C}_1 = b \mathbf{u} \tag{3} $$
where $b\in \mathbb R.$
My question
- I'm not sure if I've established the equations 1 to 3 correctly, but assuming they are, I still don't know how I can establish the system of linear equations that allows me to solve for the coordinates of $C_1$ and $C_2.$
- My hunch is that I should somehow determine the tangent points $\mathbf{P}_1$ and $\mathbf{P}_2$, but I don't see how.
- Any hints or guidance would be much appreciated here.
Your equations $(1),(2)$ do not guarantee a solution since the centres $C_i$ we are looking for may infact satisfy $C_{1,2}-P_{1,2}=\color{red}-a\frac{\mathbf n}{\|\mathbf n\|}$ or $C_1-P_1=-a\frac{\mathbf n}{\|\mathbf n\|},C_2-P_2=\color{red}+a\frac{\mathbf n}{\|\mathbf n\|}$ in case the spheres are located on opposite sides of the plane.
I would have solved the question like this: let $P:Ax+By+Cz-D=0$, so $P_{+,-}:\frac{Ax+By+Cz-D}{\sqrt{A^2+B^2+C^2}}=\pm a$ gives the equations of planes containing all the points at distance $a$ from $P$. We additionally know that the centres lie on the line $d$, so their coordinates can be found by evaluating the intersection of $d$ and $P_{+,-}$.