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Topic: geometry help
Replies: 3   Last Post: Sep 13, 2006 9:19 AM

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Vijaya Prasad Nalluri

Posts: 8
From: India
Registered: 4/10/06
Re: geometry help
Posted: Sep 12, 2006 1:14 PM
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> 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 = r-h
By Pythagoras OA^2 = ON^2 + AN^2
r^2 = (r-h)^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



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