Find point with same distance but $x$ angle from another point

Question

This is probably simple trigonometry however I've forgot how to math. How do I get the values of $X_2, Y_2$ and angle $O$ if I have the values of $X_1, Y_1$ and angle $A$?, where $A$ is the angle between line $L_1$ and $L_2$ and $O$ between $L_1$ and the $x-axis$

Answer 1

To solve for $O$ just use definition of tangent:

$$O= \arctan\left( \frac{Y_1}{X_1}\right)$$

$(X_2, Y_2)$ is obtained by rotating $(X_1, Y_1)$ by angle $A$ anti-clockwise. This can be computed by a multiplication of a rotational matrix.

$$\begin{bmatrix} X_2 \\ Y_2\end{bmatrix} = \begin{bmatrix} \cos A & - \sin A \\ \sin A & \cos A\end{bmatrix} \begin{bmatrix} X_1 \\ Y_1\end{bmatrix}$$

Answer 2

First find angle $O$: $$\tan O = \frac{Y_1}{X_1} \implies O=\arctan\left(\frac{Y_1}{X_1}\right).$$ To see the tangent, draw the line perpendicular to the $x$-axis passing through $(X_1,Y_1)$ and see the right angled triangle. Now, we can find $O+A$ in terms of $X_2$ and $Y_2$ using similar logic: $$ O+A = \arctan\left(\frac{Y_2}{X_2}\right) $$ And thus $$ A = \arctan\left(\frac{Y_2}{X_2}\right) - \arctan\left(\frac{Y_1}{X_1}\right).$$ Now, bring the expression for $O$ to the LHS and take the tangent on both sides to get $Y_2$ in terms of $X_2$. Since $(X_2,Y_2)$ is on the same circle as $(X_1,Y_1)$, then $X_1^2+Y_1^2=X_2^2+Y_2^2$. Now you have two equations and two unknowns, which should be easy (but quite tedious) to solve.