Given a circle with it's center point \(M\), the radius \(r\) and an angle \(\alpha\) of the radius line, how can one calculate the tangent line on the circle in point \(T\)? The trick is, that the radius line and the tangent line are perpendicular. This means, that their dot-product must be zero.

So, how can this schematic be derived? The intersection point \(T\) is the center point \(M\) plus the radius in the direction of the angle:

\[T = \left(\begin{array}{c} T_x\\ T_y \end{array}\right) = M + r \cdot \left(\begin{array}{c} \sin\alpha\\ \cos\alpha \end{array}\right)\]

Now we need to rotate the vector by 90° in order to get a vector that is perpendicular to the radius line:

\[\vec{v} = \left(\begin{array}{cc} 0 & 1\\ -1 & 0 \end{array}\right) \cdot \vec{MT}\]

Which leads to the line equation:

\[L(s) = T + s \cdot \frac{\vec{v}}{|\vec{v}|}\]

The implementation is therefore also quite simple:

```
function tangentLine(M, r, alpha) {
T = {
x: M.x + Math.cos(alpha) * r,
y: M.y + Math.sin(alpha) * r
}
v = {
x: T.y - M.y,
y: M.x - T.x
}
norm = Math.hypot(v.x, v.y)
v.x/= norm
v.y/= norm
return [T, v]
}
```