Difficult, but I can get you started.Call the distance between the two points, d.This can be calculated from: d = sqrt[(x2 - x1)^2 + (y2 - y1)^2]Draw the triangle containing the angle.Let t = angle. Two sides of the triangle are the radius, R,so the triangle is isosceles.Then we can say :R / sin(90 - t/2) = d / sin(t)R = d * sin(90 - t/2) / sin(t)R = d * cos(t/2) / [2 sin(t/2)cos(t/2)]R = d / [2sin(t/2)]This gives you a value for the radius.There is not enough information to say on what side ofthe line d, is the centre, so there will be two answers for this.Call the coordinates of the centre, (h,k).You will need to find two equations to act as simultaneousequations to find h and k.One springs to mind : the distances from the perimeterpoints to the centre are equal, being the radius.So, sqrt[(x1 - h)^2 + (y1 - k)^2] = sqrt[(x2 - h)^2 + (y2 - k)^2].As for the other one, I'm not sure what would be suitable.It all looks like a lot of tedious calculations.Maybe the slope formula would help :tan(t) = (m2 - m1) / (1 + m1*m2)but then you would have the slopes to deal with :m1 = (y1 - k) / (x1 - h) and m2 = (y2 - k) / (x2 - h).Good luck with that lot!!!
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Twopoint on the perimeter of circle and the angle made by them with the center...
09-07-2007, 11:15
