From: "Earl F. Glynn"
Newsgroups: borland.public.delphi.winapi
References: <3AD36D00.519ACDD@musiker.nu> <3ad3d0b3_2@dnews> <3AD3FD31.5806F33B@musiker.nu>
Subject: Re: Circle
Date: Wed, 11 Apr 2001 08:58:29 -0500
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Organization: efg's Computer Lab
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"Erik Ronström" <gitarrerik@musiker.nu> wrote in message news:3AD3FD31.5806F33B@musiker.nu...
> Yes, I knew this algorithm from school, but I didn't really know how to solve the equation between
the
> two lines (kx+m) in code. It's easy to do by hand, though.

Do you know about Kramer's rule?

> > 1.  Given points A, B, and C (not collinear)
> > 2.  Compute midpoint of AB and BC.
> > midpoint of AB = [(AX + BX)/2, (AY+BY)/2] -- just average the x and y values for the two points.

The result of this is the two midpoints.  Let's name the midpoint of AB to be D, and the midpoint of
BC to be E.

> > 3.  Compute the slopes of the lines AB and BC using the two point formula:
> > m = slope of AB = (BY-AY)/(BX-AX)  -- you'll have a special case to deal with for a vertical
line.

So you have an m1 for AB and an m2 for BC.

> > 4.  From (3) compute the slope of a perpendicular line m' = -1/m.  Remember that when the
product of
> > the slopes of two lines is -1, the lines are perpendicular?

Here you have a m1' for AB and an m2' for BC.

> I've come this far. I hope I can implement step 5 - 7, because that is what I need.
>
> > 5.  Using the point-slope formula, write the equations for the two perpendicular lines:
> > y = y1 + m'(x - x1)

Point D and slope m1':  y = DY + m1'(x - DX)
Point E and slope m2':  y = EY + m2'(x - EX)

DX, DY, EX, EY, m1' and m2' are just numbers at this point.  You have two equations in x and y.

> > 6.  The solution of the two equations in two unknowns from (5) will give you the center of the
> > circle, (x,y).

To keep this simple, let me rename the constants:

Do you know about Cramer's Rule two equations in two unknowns.  It's explained here.
http://www.iln.net/html_p/c/53785/53786/53801/53873/54176.asp

To use Cramer's first you must rearrange your equations:

m1'*x - y = m1'*DX - DY
m2'x - y = m2'*EX - EY

Instead of Cramer's rule, you could just take the negative of one of the equations and add it to the
other equation.  This would let you solve for x.  Once you have x, you could solve for y using one
of the original equations.

An iterative solution is not necessary.  Everything is well defined for a direct solution.
--
efg     efg2@efg2.com     Earl F. Glynn, Overland Park, KS  USA

efg's Computer Lab:  http://www.efg2.com/Lab
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