While knowing the radii of the cylinder and the points in the center of the top and bottom plane, how would you find the equation of the points on the surface of the cylinder? The slices of the cylinder are circular, but the cylinder itself might be rotated in 3 dimensions.
I don't know if I explained that perfectly, so here's an example cylinder of what I mean.
Let us treat the particular case of a cylinder of unit height $OC$ and radius $r$. The general case will be explained at the end.
As $v=\vec{OC}=\pmatrix{a\\b\\c}$ is such that $\|v\|^2=v^Tv=1$, it is known that the orthogonal projection operator $\cal{P}$ : $M \to N$ onto the axis $OC$ is given by matrix
$$\cal{P}=vv^T=\pmatrix{a^2&ab&ac\\ab&b^2&bc\\ac&bc&c^2}$$
Therefore, $\vec{ON}=vv^T \vec{OM}.$
As a consequence : $\vec{NM}=\vec{OM}-\vec{ON}=Id(\vec{OM})-vv^T (\vec{OM})$
$$\vec{NM}=(Id-vv^T) \vec{OM}$$
It remains to express that $\|\vec{NM}\|^2=r^2$ :
$$((Id-vv^T) \vec{OM})^T(Id-vv^T) \vec{OM}$$
$$(\vec{OM})^T \underbrace{(Id-vv^T)^T(Id-vv^T)}_{(Id-vv^T)} \vec{OM}=r^2$$
$$(x \ y \ z) \pmatrix{1-a^2&-ab&-ac\\-ab&1-b^2&-bc\\-ac&-bc&1-c^2} \pmatrix{x\\y\\z}=r^2\tag{1}$$
Now, if the length of $OC$ is $L$ and the basis' center is in $(x_0,y_0,z_0)$ instead of $(0,0,0)$, it remains to do an enlargment followed by a translation, i.e., replace coordinates $x,y,z$ in (1) by
$$\begin{cases}X=\frac{x-x_0}{L}\\Y=\frac{y-y_0}{L}\\Z=\frac{z-z_0}{L}\end{cases}$$
and replace $r$ by $\frac{R}{L}$.