How to calculate the coordinates on an arc that divides it into five equal parts? I have a circle sector for which I know:
The coordinates of the three points $A, B$, and $S$, and the radius of circle $S$.

I've worked on something similar here, but it was midpoint coordinates.
Now I need $4$ points to "divide" the arc into five equal parts. How do I calculate it, please?
Thank you.
PS. I calculated the midpoint coordinates using a vector (it's written here)
 A: First, determine the angle of the entire arc, i.e., $\alpha = \measuredangle ASB$. For this, calculate the perpendicular, i.e., shortest, distance from $A$ to $SB$. This is given by the formula for the distance to a line defined by two points, where with $r = \sqrt{(y_2 - y_3)^2 + (x_2 - x_3)^2}$ is the radius of the circle, we get:
$$p = \frac{\left|(y_2 - y_3)x_1 - (x_2 - x_3)y_1 + x_2y_3 - y_2x_3\right|}{r} \tag{1}\label{eq1A}$$
If the point where this perpendicular meets $SB$ is $C$, the $\triangle ACS$ is right-angled at $C$, so we then get
$$\sin(\alpha) = \frac{p}{r} \implies \alpha = \arcsin\left(\frac{p}{r}\right) \tag{2}\label{eq2A}$$
Note, however, the determined value of $\alpha$ assumes $\measuredangle ASB \le \frac{\pi}{2}$. However, with $SA$ being perpendicular to $SB$, there are $2$ points where $A$ can be which gives the same value of $\alpha$ and, otherwise, there are $4$ possible points, as indicated in the diagram below.

Note $\measuredangle A_1SB = \alpha$, $\measuredangle A_2SB = \pi - \alpha$, $\measuredangle A_3SB = \pi + \alpha$ and $\measuredangle A_4SB = 2\pi - \alpha$.  If there are bounds or other conditions allowing you to already know which one is correct, e.g., $\measuredangle ASB \le \frac{\pi}{2}$ so $A$ is $A_1$ and the angle is $\alpha$, then you can just use that angle. Otherwise, there are several ways to determine which point, and thus angle, is the correct one. Here is a relatively simple method to use.
The vector $\mathbf{v_1}$ going from $S$ to $B$ is
$$\mathbf{v_1} = (x_2 - x_3, y_2 - y_3) = (x_4, y_4) \tag{3}\label{eq3A}$$
Let $\theta_i$ for $1 \le i \le 4$ be each of the $4$ possible values of $\measuredangle A_{i}SB$ given above, and the vector from $S$ to $A_{i}$ be $\mathbf{v_{2,i}}$. Then the rotation matrix formula gives
$$\mathbf{v_{2,i}} = (x_4\cos(\theta_i) - y_4\sin(\theta_i), x_4\sin(\theta_i) + y_4\cos(\theta_i)) = (x_{5,i}, y_{5,i}) \tag{4}\label{eq4A}$$
which means
$$A_i = (x_3 + x_{5,i}, y_3 + y_{5,i}) = (x_{6,i}, y_{6,i}) \tag{5}\label{eq5A}$$
Due to errors, usually quite small, in the determined trigonometric values and rounding errors in the calculations, it's likely none of the $A_i$ values will match those of $A$ exactly. You can use something like checking the absolute values of the $x$ and $y$ co-ordinates being very close or, alternatively, determine the smallest distance using
$$d_i = \sqrt{(x_{6,i} - x_1)^2 + (y_{6,i} - y_1)^2} \tag{6}\label{eq6A}$$
Once the appropriate $\theta_i$ angle is determined, the first arc point $D$ to determine is a rotation of $\mathbf{v_1}$ by an angle of $\beta = \frac{\theta_{i}}{5}$, so let the vector from $S$ to $D$ be $\mathbf{v_3}$. The rotation formula then gives
$$\mathbf{v_{3}} = (x_4\cos(\beta) - y_4\sin(\beta), x_4\sin(\beta) + y_4\cos(\beta)) = (x_{7}, y_{7}) \tag{7}\label{eq7A}$$
which means
$$D = (x_3 + x_{7}, y_3 + y_{7}) \tag{8}\label{eq8A}$$
You can use a similar procedure for the other $3$ arc points to determine.
A: Write $s=x_3+i y_3$, compute the radius $r=((x_2-x_3)^2+(y_2-y_3)^2)^{1/2}$, compute $\theta_0 = \text{atan} \frac{y_2-y_3}{x_2-x_3}$, and $\theta_1=\text{atan} \frac{y_1-y_3}{x_1-x_3}.$
Now $b=s+re^{i\theta_0}$ and $a=s+re^{i\theta_1}.$
The four points are $s+re^{i[\theta_0 + \frac{k}{5} (\theta_1-\theta_0)]},\quad k\in\{1,2,3,4\}$
