Image for Reference

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.

  • $\begingroup$ Do you confirm the initial $3$ ? $\endgroup$
    – user65203
    Oct 28, 2019 at 13:33
  • $\begingroup$ Would probably be helpful to show us the original geometric setup. $\endgroup$
    – user65203
    Oct 28, 2019 at 13:35
  • $\begingroup$ Is d=distance from $E_{init}$ to $E$ ? Are $(X_1,Y_1)$ the cordinates of $E_{init}$ and $(X_2,Y_2)$ the coordinates of $E$ ? In this case, I don't understand the angle $\alpha$ in your figure if I follow your definition $\alpha =\left( \arccos{\frac {Y_2-Y_1}{d}}+\psi \right)$ (are the parentheses well placed ?) : the arc shouldn't go till line $E_{init}C_{\ell}$... Besides, coordinates $(X_3,Y_3)$ correspond to which point ? $\endgroup$
    – Jean Marie
    Oct 28, 2019 at 21:36
  • $\begingroup$ @JeanMarie I have added more details into the question regarding the coordinates. $\endgroup$
    – Harsha
    Oct 29, 2019 at 7:09
  • $\begingroup$ @JeanMarie Yes the parenthesis placed are correct for angle $\alpha$. The idea is to move the point $E_{init}$ such that the new found $d$ and $\alpha$ when used in applying Al-Kashi's theorem to the triangle $C_lE_{init}C_r$ should give $R_{E_{init}}$ $=$ $R_{min}$ $\endgroup$
    – Harsha
    Oct 29, 2019 at 7:16


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