Solve a system of quadratic and trigonometrc equations

Question

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.

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

$$\cos(\alpha-\psi)=\sin\psi$$

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

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