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I'm defining an arc by calling a function like this: arc(x, y, radius*2, radius*2, start, end);

x: center x.
y: center y.
radius*2: width.
radius*2: height.
start: start angle in radians or degrees.
end: end angle in radians or degrees.

And now I need to "close" the arc by drawing two lines, both from the center of the arc, to the start and end point of the arc.

How do I find the start and end points?

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  • $\begingroup$ You know the parametric equations for a circle? $\endgroup$ – J. M. is a poor mathematician Aug 20 '12 at 13:51
  • $\begingroup$ No I don't think so. $\endgroup$ – 01AutoMonkey Aug 20 '12 at 14:01
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    $\begingroup$ Well... you have some reading to do then!. If you get stuck come back with your sticking point. $\endgroup$ – rschwieb Aug 20 '12 at 14:27
  • $\begingroup$ If you have different values for width and height, you have an ellipse, not a circle. Why is radius*2 repeated in the argument list? $\endgroup$ – Ross Millikan Aug 20 '12 at 14:33
  • $\begingroup$ Are you working in GeoGebra? $\endgroup$ – Sigur Aug 20 '12 at 14:45
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Given start angles and end angles you get

sX = offsetX + radius * cos(startAngle * PI/180)
sY = offsetY + radius * sin(startAngle * PI/180)
eX = offsetX + radius * cos(endAngle * PI/180)
eY = offsetY + radius * sin(endAngle * PI/180)

Multiplying an angle by PI/180 converts degrees to radians.

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After reading http://en.wikipedia.org/wiki/Equation_of_a_circle#Equations as suggested by @rschwieb I came up with the following:

s_x = x+radius*Math.cos(startAngle*Math.PI);
s_y = y+radius*Math.sin(startAngle*Math.PI);
e_x = x+radius*Math.cos(endAngle*Math.PI);
e_y = y+radius*Math.sin(endAngle*Math.PI);
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  • $\begingroup$ If your angles are in radians, you do not need the Math.PI factor. If your angles are in degrees, then you will need a factor of Math.PI/180 instead. $\endgroup$ – robjohn Aug 22 '12 at 19:16

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