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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?

enter image description here

Thank you

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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 $$S+ \text{radius}\cdot \vec u$$

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  • $\begingroup$ Yes, it's a midpoint. I'm sorry, but I don't understand. I can calculate the midpoint between points A and B, on an imaginary line between these points. But how will it help me? I don't know. $\endgroup$ – yoda666 Jun 12 '20 at 21:26
  • $\begingroup$ You can get the unit vector $\vec u$ I mention in my answer, so you have a parametric equation for the arc midpoint $\endgroup$ – Bernard Jun 12 '20 at 21:28
  • $\begingroup$ Sorry for the late reply. I already understood that, that's exactly what I needed. Thank you. $\endgroup$ – yoda666 Jun 23 '20 at 15:17
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Claim: If you draw a line from $S$ to the midpoint of the arc (call it $M_1=(x_4, y_4)$). And draw a line from $A$ to $B$ ($\overline{AB}$). The two lines ($\overline{SM_1}$ and $\overline{AB}$) will intersect at that midpoint of the line from $\overline{AB}$ (call that point $M_2$).

Pf: Draw triangles $\triangle M_2SA$ and $\triangle M_2SB$. They are congruent by $SAS$ ($M_2S = M_2S$ and $\angle M_2SA \cong \angle M_2SB$ and $SA = SB$) so $AM_2 = BM_2$.

.....

we know:

  1. $M_2$ is the midpoint of $\overline{AB}$

  2. $S, M_1$ and $M_2$ are all on the same line

  3. $SA= SB =SM_2$

So we have three sets of equations. Us them to solve for $x_4, y_4$.

  1. Equation of midpoint

so $M_2 = (\frac {x_1+x_2}2, \frac {y_1+y_2}2)$.

  1. Equation of slope

$\frac {y_4-y_3}{x_4-x_3} = \frac {\frac {y_1+y_2}2- y_3}{\frac {x_1+x_2}2-x_3} = \frac {y_4-\frac {y_1+y_2}2}{x_4-\frac {x_1+x_2}2}$

  1. distance formula

$\sqrt{(x_1-x_3)^2 + (y_1-y_3)^2} = \sqrt{(x_4-x_3)^2 + (y_4-y_3)^2} = \sqrt{(x_2-x_3)^2 + (y_2-y_3)^2}$

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  • $\begingroup$ Sorry for the late reply. Your solution may seem simple. However, it is easier to calculate this using the vector in the answer above. You just need to know how to work with vectors. Thank you anyway. $\endgroup$ – yoda666 Jun 23 '20 at 15:21

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