Finding the angle between the $x$-axis and a tangent drawn to a circle from the origin

Question

Suppose that a circle $\mathcal{C}$ centred at $(x_0,y_0)$ with radius $r$ is at a distance $L$ from the origin such that $x_0^2 + y_0^2 > r^2$, i.e. distance from the origin to $(x_0, y_0)$ is $L + r$. Can we then express the angles $\alpha_1,\alpha_2$ between the $x$-axis and the tangents of $\mathcal{C}$ which intersect the origin in terms of $L$ and $r$ in a coordinate free manner, which in this case means that we do not use vector addition to explicitly compute the intersections $T_1,T_2$ of $\mathcal{C}$ and the two tangents then compute $\alpha_1,\alpha_2$ by first finding the lengths of the sides of the right triangles $OT_1P_1$ and $OT_2P_2$ where $P_1,P_2$ are the projections of $T_1,T_2$ onto $x$-axis?

Answer 1

Let the line joining the origin to the center of the circle be inclined at an angle $\theta$ from the $x$-axis.

From trigonometry, $\tan{\theta}=\frac{y_0}{x_0}$

The angles $\alpha_1$ and $\alpha_2$ must be equally spaced from $\theta$, i.e., $\alpha_1=\theta-\omega$ and $\alpha_2=\theta+\omega$

From trigonometry, $\sin{\omega}=\frac{r}{L+r}$

The required angles are $\alpha_1=\arctan\left(\frac{y_0}{x_0}\right)-\arcsin\left(\frac{r}{L+r}\right)$ and $\alpha_2=\arctan\left(\frac{y_0}{x_0}\right)+\arcsin\left(\frac{r}{L+r}\right)$