I have a condition

$$3R^2-d^2+2dR\cos\alpha = 0$$

along with the following equations

$$d = \sqrt{(X_1-X_2)^2+(Y_1-Y_2)^2}$$ $$\alpha = \arccos{\frac {Y_2-Y_1}{d}}+\psi$$ $$Y_2=Y_1-(X_1 - X_2)\tan\psi $$

The values $ X_1, Y_1, \psi, R$ 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.

  • $\begingroup$ You want to solve this System of equations for $X_2,Y_2,d$? $\endgroup$ Oct 22, 2019 at 14:03
  • $\begingroup$ Yes, since $Y_2, d$ can be expressed in terms of $X_2$, obtaining the solution for $X_2$ is the goal. $\endgroup$
    – Harsha
    Oct 22, 2019 at 14:07
  • 1
    $\begingroup$ This is the Output by Mathematica, i hope this will help you: $$\left(\left(0=d-\sqrt{d^2}\land \psi =\alpha -\frac{\pi }{2}\land \text{X2}=\frac{-\sqrt{d^2 \left(\cot ^2(\alpha )+1\right)}+\text{X1} \cot ^2(\alpha )+\text{X1}}{\cot ^2(\alpha )+1}\right)\lor \left(0=d-\sqrt{d^2}\land \psi =\alpha -\frac{\pi }{2}\land \text{X2}=\frac{\sqrt{d^2 \left(\cot ^2(\alpha )+1\right)}+\text{X1} \cot ^2(\alpha )+\text{X1}}{\cot ^2(\alpha )+1}\right)\right)\land \text{Y1}=\text{X1} \tan (\psi )-\text{X2} \tan (\psi )+\text{Y2}$$ $\endgroup$ Oct 22, 2019 at 14:44
  • $\begingroup$ @Dr.SonnhardGraubner I did not really understand the solution, also there were a few mistakes in the question posted, I have done the corrections now, Can you please check again? I have posted the question again with more details link $\endgroup$
    – Harsha
    Oct 29, 2019 at 8:09

1 Answer 1


Let $x=X_2-X_1$ and $y=Y_2-Y_1$. From the givens,

$$ y = x \tan\psi, \>\>\>\>\> d= x\frac{\tan \psi}{\cos(\alpha-\psi)}$$

Plug above into $d = \sqrt{x^2+y^2}$ to get


which has the solution $\alpha=\frac \pi2$, or $\alpha = 2\psi -\frac\pi2$. In turn, $\cos\alpha=0$, or $\cos\alpha=\sin2\psi$.

For $\cos\alpha=\sin2\psi$, we have

$$3R^2-d^2+2dR\sin2\psi = 0$$

which leads to

$$d= R\left(\sin2\psi+\sqrt{\sin^22\psi+3}\right)$$

As a result,

$$x = d \frac{\cos(\alpha-\psi)}{\tan\psi}= d \cos\psi= R\cos\psi\left(\sin2\psi+\sqrt{\sin^22\psi+3}\right)$$

The other solution, corresponding to $\cos\alpha=0$, is $d = \sqrt3 R$, which yields

$$x= \sqrt3 R\cos\psi$$

  • $\begingroup$ I have posted the question again with more details, can you please check. There were mistakes in the question i posted here. Link for the new question $\endgroup$
    – Harsha
    Oct 29, 2019 at 8:11

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