I have been given a task to write head up display screen for the drone we are working on. The work has come up to the showing the roll degrees, which the mathematical form of the problem is as below:
Problem: Let's say we have a circle with the equation of $(x - c_0)^2 + (y-c_1)^2 = r$. What is the equation of tangent line touches the circle at $\theta$ degree?
Thanks in advance.
1) Parametric representation of the tangent
As vector $\vec{T}:=\binom{r \sin(\theta)}{-r \cos(\theta)}$ is orthogonal to $\vec{OP}$ (their dot product is $0$), we can write that a general point $M=\binom{x}{y}$ belonging to the tangent line is such that $\vec{PM}=t\vec{T} \ \Leftrightarrow \ M=P+t\vec{T}$, otherwise said, because $P=\binom{c_0+r \cos(\theta)}{c_1+r \sin(\theta)}$:
$$\begin{cases}x(t)&=&c_0+r \cos(\theta)+tr\sin(\theta)\\y(t)&=&c_1+r \sin(\theta)-t r \cos(\theta)\end{cases}$$
If we desire for example the coordinates of point $F$, it suffices to find the values of $t$ to which it is associated. It is such that the $y$ oridnate is $0$, giving $t=t_M=\frac{c_1+r \sin(\theta)}{r \cos(\theta)}.$
2) Cartesian equation of the tangent
For this issue and other issues of the same type, here is a more general result :
The tangent line in $(x_0,y_0)$ to a conic (circle, ellipse, parabola, hyperbola) with equation:
$$\tag{1}ax^2+by^2+2cxy+2dx+2ey+f=0$$
is
$$\tag{2}axx_0+byy_0+c(xy_0+x_0y)+d(x+x_0)+e(y+y_0)+f=0$$
It suffices for you to apply (2) in the case of the conic with equation: $$(x - c_0)^2 + (y-c_1)^2 = r^2 \Longleftrightarrow x^2 + y^2 -2 c_0x -2 c_1y + (c_0^2+c_1^2-r^2)=0 \ \text{at point}$$
$$(x_0,y_0)=(c_0+r \cos(\theta),c_1+r \sin(\theta)).$$
There remains a last computation. Up to you...