I have shape for which I know:
Coordinates of three points (A, B, S) and radius.
How do I calculate the coordinates of the center (midpoint) of the arc between points A and B, please?
Thank you
Calculate the coordinates of the center (midpoint) of the arc
geometry
Related Solutions
Don't write down too many equations to be solved, but produce the desired center ${\bf c}=(a,b)$ in a forward movement instead. Let ${\bf z}_i=(x_i,y_i)$ $\ (i=0,1)$ be the two given points, put $\epsilon:=1$ if the arc should go from ${\bf z}_0$ to ${\bf z}_1$ counterclockwise, and put $\epsilon:=-1$ otherwise.
Next, let $d:=|{\bf z_1}-{\bf z_0}|=\sqrt{(x_1-x_0)^2+(y_1-y_0)^2}$ be the distance and ${\bf m}:=\bigl({x_0+x_1 \over2}, {y_0+y_1\over2}\bigr)$ be the midpoint of ${\bf z_0}$ and ${\bf z_1}$. Then
$${\bf n}:=(u,v):=\Bigl({x_1-x_0\over d},{y_1-y_0\over d}\Bigr)$$
is the unit normal in the direction ${\bf z_1}-{\bf z}_0$, and ${\bf n}^*:=(-v,u)$ is the unit vector you get by rotating ${\bf n}$ counterclockwise by $90^\circ$.
Given $r>0$ the center ${\bf c}$ has a distance $h:=\sqrt{r^2 -d^2/4}$ from ${\bf m}$, and the given $\epsilon$ together with ${\bf n}^*$ tell us in which direction we should go. In vectorial notation the center is given by
$${\bf c}\ =\ {\bf m}+\epsilon\ h\ {\bf n}^*\ ,$$
so that coordinate-wise we get
$$a={x_0+x_1 \over2}-\epsilon\ h\ v, \qquad b={y_0+y_1 \over2}+\epsilon\ h\ u\ .$$
WLOG, we can let the circle be centered at O(0, 0) with radius = r.
Therefore, the equation of the circle is $x^2 + y^2 = r^2$
M(p, q) is point on this circle implies $p^2 + q^2 = r^2$ ……… (1)
By midpoint formula, $N(r, s) = N(\dfrac {x_1 + x_2}{2}, \dfrac {y_1+ y_2}{2})$
N(r, s) is a point on OK, the line perpendicular to $P_1P_2$. By two-point form, the equation of OK is
$y = \dfrac {y_1 + y_2}{x_1 + x_2}x$
M(p, q) is also a point on OK. Thus,
$q = \dfrac {y_1 + y_2}{x_1 + x_2}p$ ………. (2)
Solving (1) and (2) will give you $p = ± r \dfrac {x_1 + x_2}{\sqrt{(x_1 + x_2)^2 + (y_1 + y_2)^2}}$
The ‘±’ provides two sets of answers for (p of M) and (p’ for M’) as shown.
The corresponding values of q can be found via (2).
Selecting the correct M(p, q) is another story.
Best Answer
The ‘centre’ of the arc is its midpoint. It's fairly easy: the line through $S $ and this midpoint is the bissectrix of the angle $\widehat{ASB}$, and as you have an isosceles triangle, it is also the median through $S$ of the triangle.
Therefore, once you have determined the midpoint $I$ of the segment $[AB]$, the unit directing vector of the median is $\vec u=\frac{\overrightarrow{SI}}{\|\overrightarrow{SI}\|}$, and the midpoint of the arc is simply the point $$S+ \text{radius}\cdot \vec u$$