Ahne gets you almost all the way there mathematically. You now know the
distance from the center of the chord to the center of the circle. You
don't yet know the actual coordinates of the center. Apply the Pythagorean
theorem a couple more times and solve the simultaneous equations and you got
it.
From Ahne:
(r^2-h^2)^0.5= d, distance from x3,y3 to x4,y4 (the circle center) we need
to find x4,y4
d^2=(x4-x3)^2 + (y4-y3)^2 two unknowns (x4,y4), one equation
we also know:
r^2=(x4-x1)^2 + (y4-y1)^2 two equations, two unknowns, a little
algebra and it's solved
Whew, it is much easier just with a compass and a ruler!!!
-----Original Message-----
From: Ahne Oosterhof [mailto:[log in to unmask]]
Sent: January 24, 2002 5:48 PM
To: [log in to unmask]
Subject: Re: [TN] center of a arc
To construct this on paper is real easy. Draw a straight line between two
points on the arc (called a chord). Then use your compass and draw two
circles each with the same radius as the arc, centered on the end points of
the chord. Where these two circles intersect each other is the center of the
arc. (You can choose which one of the two intersections is the correct one.)
To calculate it with geometry is a little more involved:
You know the two points on the arc: x1,y1 and x2,y2. The centerpoint of the
chord is (x1-x2)/2, (y1-y2)/2, or x3,y3. Half the length of the chord is
h=((x3-x1)^2 +(y3-y1)^2)^0.5. This is one side of the right triangle, of
which the hypotenuse is the arc radius and the third side is the distance
from the center of the chord to the center of the arc. Use Pythagorean
theorem once more: (r^2-h^2)^0.5, and you end up where you want to be.
Regards,
Ahne.
-----Original Message-----
From: TechNet [mailto:[log in to unmask]] On Behalf Of Matt Eubank
Sent: Thursday, January 24, 2002 07:30
To: [log in to unmask]
Subject: [TN] center of a arc
Need some help.
What is the equation for finding the center of a arc given the radius and 2
points on the arc.
Thanks
Matt
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