Reposting again since I had mentioned the details incorrectly
I have a condition
$$3R^2-d^2+2dR\cos\alpha = 0$$
along with the following equations
$$d = \sqrt{(X_1-X_2)^2+(Y_1-Y_2)^2}$$ $$\alpha = \arccos{\frac {Y_2-Y_1}{d}}+\psi$$ $$Y_2=Y_3-(X_3 - X_2)\tan\psi $$
The values $ X_1, Y_1, \psi, R, X_3, Y_3 $ are known.
The resolution of the first condition should lead to a second order equation for $X_2$. The solution obtained for $X_2$ is of interest.
Tried solving it with little success, someone help me out.
Edit: More details
$(X_1,Y_1)$ are the coordinates of the point $C_l$
$(X_3,Y_3)$ are the coordinates of the point $E_{init}$
$(X_2,Y_2)$ are the coordinates of the new position of point $E_{init}$
$R = R_{min}$
$d$ is the distance between $C_l$ and the new position of $E_{init}$
Distance between $C_l$ and $C_r$ is $R_{min} + R_{E_{init}}$
The idea behind the above is to find out the new position for $E_{init}$ shown in the image above when $R_{E_{init}}$ is less than $R_{min}$ in order to make $R_{E_{init}}$ same as $R_{min}$. Once the new position of the $E_{init}$ is calculated the new position of $C_r$ is found, therefore reducing the distance between $C_l$ and $C_r$ to $2*R_{min}$.
The conditions mentioned are obtained from a journal.
