
Re: geometry help
Posted:
Sep 12, 2006 1:14 PM


> Hi, > I need some help, please. > I have : >  an circle arc AB >  x1,y1,x2,y2  coordonates of A(x1,y1) and > nd B(x2,y2). >  h  height of arc >  d  lenght of line segment AB > I want to know the coordonates of the circle. > > Thanks a lot. > PS : I don't speek english very good.
Hi Dane,
Let r be the radius of the circle Drop a perpendicular ON upon AB from the center of the circle O Triangle ONA is right angled at N with AN = d/2, OA = r, and ON = rh By Pythagoras OA^2 = ON^2 + AN^2 r^2 = (rh)^2 + (d^2)/4 2rh = h^2 + (d^2)/4 r = (h/2) + ((d^2)/8h) Assume the equation to the circle to be x^2 + y^2 + 2gx + 2fy + c = 0 ,,,,,,(*) So radius squared is g^2 + f^2  c = (h/2) + ((d^2)/8h) .........(1) The circle passes through A(x1,y1) and B(x2,y2) Therefore x1^2 + y1^2 + 2gx1 + 2fy1 + c = 0 x2^2 + y2^2 + 2gx2 + 2fy2 + c = 0 which may be rewritten as 2gx1 + 2fy1 + c = (x1^2 + y1^2) .........(2) 2gx2 + 2fy2 + c = (x2^2 + y2^2) .........(3) Solve equations (1), (2), and (3) for g, f, c and substitute their values in (*)to get the equation of the circle whose radius r is (h/2) + ((d^2)/8h) and center is ( g, f) [e.g. : Let A =(1,0), B =(5,0), (so d = 4) and h = 2 From (2) and (3) 1 + 2g + c = 0, 25 + 10g + c = 0, which give g =  3, c = 5 also 9 + f^2  5 = 1 + 4 from (1) giving f = 1, radius = sq rt of 5 center = (3,1) eq to the circle is x^2 + y^2 6x + 2y +5 = 0]
Vijayaprasad
P.S.:Note that d^2 = (x1  x2)^2 + (y1  y2)^2 So it suffices if we know coordts of A, B and value of h

