Please refer the image.
How to figure out the angle $α = \widehat{C_{l}E_{init}C_{r}}$ with only the following information
$d_{ClCr} = R_{E_{l}min} + R_{E_{init}r}$
The lines $E_{init}C_{r}$ and $E_{init}D$ are perpendicular.
$R_{E_{init}r}$ and $C_{r}$ are unknown and will be deduced with the help of $α$.
$C_{l}$, $R_{El_{min}}$, $ψ$, $d_{C_{l}E_{init}}$ are known.
Let's call $R_{E_{\text{init}r}}$ simply $R_1$ and the other $R$ simply $R_2$; let's call $E_\text{init}$ simply $E_1$. To be consistent, we then relabel $C_r$ as $C_1$, and $C_\ell$ as $C_2$. So the idea is that we rotate $E_1$ about $C_1$ until it reaches a point $E_2$ on the line $C_1C_2$. Then we rotate $E_2$ about $C_2$ until it reaches $E$.
$R_2$ is known. $C_1C_2=d$ is known, so $R_1=d-R_2$.
We are told that $E_1D_1$ is perpendicular to the radius $C_1E_1$. In other words it is a tangent to the circle centre $C_1$. So the angle between $C_2E_1$ and the tangent at $E_1$ is $\psi$, which is known.
Hence $\angle C_2E_1C_1=\alpha=90^o+\psi$. That is apparently all that the OP requires.
But pressing on a little: $C_2E_1=d_2$ is known. But we need to know $\angle C_1C_2E_1$. Then the triangle $C_1C_2E_1$ is determined and we can find $\angle EC_2E_2$ and $\angle E_1CE_2$.
Addid later.
@HarshaUmapath You now say $R_1$ is unknown. That does not affect the angle $\alpha$, but it does mean that the configuration is not determined, because there is no way of knowing the position of $C_1$ on the line $E_1C_1$.