# Solution for a system of quadratic and trigonometric equations

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.

• Do you confirm the initial $3$ ?
– user65203
Oct 28, 2019 at 13:33
• Would probably be helpful to show us the original geometric setup.
– user65203
Oct 28, 2019 at 13:35
• 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 ? Oct 28, 2019 at 21:36
• @JeanMarie I have added more details into the question regarding the coordinates. Oct 29, 2019 at 7:09
• @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 $R_{E_{init}}$ $=$ $R_{min}$ Oct 29, 2019 at 7:16