Conference rusure::math

    Two points by themselves do not uniquely determine a circle.  Any
    point on the perpendicular bisector of the segment joining the two
    points could be used as the center.
    
    Dan

    what about the circle whose radius is the distance of the two points,
    and the two points are on the circle.
    
    Concider the figure

              (x2,y2)
                 .
                /
               /
              /
             /         dy = y2 - y1
            /          dx = x2 - x1
           /           r = sqrt( sqr(dy) + sqr(dx) )
          /
         .             For this example I am interested for a circle
      (x1,y1)          whose radius =r (given above) and the two points
                       (x1,y1) and (x2,y2) are on this circle. Also
                       there are only two possible circles. I am interested
                       on the one whose center is to the right of this
                       figure. Given all these can I find the center
    of this circle?
    
    /k
    
    
    

    What I was trying to say was given that I know (x1,y1) and (x2,y2)
    and the distance between these two points r what are the two points
    (denoted by a * in the figure bellow), where the two circles, one 
    with center at (x1,y1) and radius = r and the other at (x2,y2) with 
    same radius = r, intersect?
    
       
               (x2,y2)
                  .
    *            /
                /
               /
              /
             /            
            /             *
           /
          .
       (x1,y1)
    

    Let d^2 = (x2-x1)^2 + (y2-y1)^2, and h = sqrt(4*r^2/d^2-1)
    Then

	x0 = 0.5*((x1+x2) +/- (y2-y1)*h)
	y0 = 0.5*((y1+y2) -/+ (x2-x1)*h)

    - Jim

    re. .-1
    
    Thanks Jim for your help. I have implemented your method and it
    works. I have one more question on the same subject.
    
    Now that I know the center of this new circle which is at (x0, y0)
    can I find the angles (in degrees) of the (x1,y1) and (x2,y2) points
    so I can draw an arc between them. This arc will be a portion of the
    circle whose center is the point (x0,y0) and radius=r and the 
    starting angle is at the point (x1,y1) and the ending angle is at
    the point (x2,y2).

/* This is extracted from an arc routine which generates ReGIS output.
 * It computes the angles of the segments from the center to the start
 * and from the center to the end point of the arc, using the arctangent
 * function.  It then takes the difference of the two angles and
 * converts to integral degrees for the benefit of ReGIS. 
*/
#define PI 3.1415926536
#define TWOPI (2 * PI)
arc (x0,y0,x1,y1,x2,y2)
double x0,y0,x1,y1,x2,y2;
{
  int ang; double theta;
  theta = atan2(y2-y0, x2-x0) - atan2(y1-y0, x1-x0);
  if (theta < 0.0) theta += TWOPI;
  ang = (360./TWOPI) * theta + 0.5;
  printf("P(B)[%d,%d]C(A%d)[%d,%d]P(E)",(int)x0,(int)y0,ang,(int)x1,(int)y1);
}

    Re .5:
    
    This appears to be an equilateral traingle.
    
    Therefore the angle(s) of any such arc(s) will always be 60�.  (that
    is, 60 degrees for those without multi-national character sets.)
    
    
    --  Barry

    re. .-1
    
    Yes, I am aware that the triangle is an equilateral one but my question
    was with respect to the new circle centered at (x0,y0). Once again
    at what degree does the arc, from the point (x1,y1) to the point
    (x2,y2), starts and ends respectively? The starting point is the
    most important since as you suggested the ending point may be equal
    to the starting point plus 60 degrees.
    
    /Kostas
    
    

This seems like a logical topic to append this to:

If I have three points in a cartesian coordinate system, how can I find the
coordinates of the center of the circle passing through those three points?

I should be able to derive this, but it has been a l-o-n-g time since I studied
this, and I could use the formula for some woodworking I'm doing.

Thanks,
Pete

	A segment between two of the points would be a chord
	of the circle.  The perpendicular bisectors of any two
	of the chords meet at the center of the circle.  (If
	the perpendicular bisectors are parallel, then the
	points are collinear.  You can think of the parallel
	bisectors as "meeting" at the center of a "circle at
	infinity".)

	Does that give you enough to derive a formula?  Or at
	least to find the center on the wood?

	Dan

re: -.1

Yes, that's a start and I'll start working with it - thanks!  And I'm still
open to a short formula!

    if your 3 points are at (x0,y0), (x1,y1) and (x2,y2) then the line
    segments joining them are also chords of the circle passing through
    the 3 points.  The equation of one chord is:
    
    			    (y1-y0)
    		y - y0  =   ------- (x - x0)
    			    (x1-x0)
    
    Which simplifies to y = mx + b, for constants b,m.  I don't think
    you'll need to calculate b.  m is just (y1-y0)/(x1-x0).
    
    The equation of the line perpendicular to "y = mx + b" at (x',y') is:
    
    			     -1
    		y - y'  =   ---- (x - x')
    			      m
    
    So select (x',y') as the midpoint of the line (x0,y0) to (x1,y1) and
    grind out the above equation.  Choose (x",y") as the midpoint of the
    line (x1,y1) to (x2,y2) and grind out the above formula again ("m" will
    be different this time, (y2-y1)/(x2-x1), else points are collinear).
    
    This gives you two equations in two unknowns.  The solved value for
    (x,y) will be the coordinates of the center of the circle.
    
    If you have a symbolic math package it could give you the formula directly.
    
      John

OK, here's what I figure I have to do, given 2 chords

For the first chord, find the midpoint [(x1-x2)/2, (y1-y2)/2]; this is where
the perpendicular bisector 'starts' ...

Find the slope of the chord ...

Take the negative reciprocal of that slope ... this is the slope of the
perpendicular bisector ...

Use the point-slope equation y-y1=m(x-x1) to find the equation of the
perpendicular bisector ...

Do all of the above the the second chord, too ...

Set the two equations equal and solve for "x", giving the x-coordinate of the
circle's center ...

Use the distance equation to find the radius r=((x2-x1)�-(y2-y1)�)**� ...

Whew!

Is there a more direct approach?

Pete

>> Is there a more direct approach?

	Using your pencil and a yardstick, lightly draw in
	two chords and their perpendicular bisectors on the
	wood.  Where the bisectors meet is the center of the
	circle.

	:-)
	Dan

As it turns out, for my purposes -.1 is indeed the most reasonable approach.

> As it turns out, for my purposes -.1 is indeed the most reasonable approach.
    
    But don't you at least feel better, having solved the math problem?
    
    Short of coding up a program (or fetching one from netlib :-) I don't 
    think there is a "simpler way".
    
      John

You would do well to have a compass to help make the perpendicular 
bisectors, but that's probably less effort than computing an exact solution:
Assuming the 3 points are (x0,y0), x1,y1), (x2, y2), the exact solution for 
the center (xc, yc) is:

            2        2           2        2        2     2        2        2
xc = 1/2 (x0  y1 - x0  y2 - y0 x1  + y0 x2  - y1 x2  + x1  y2 - y0  y2 + y0  y1

         2           2        2        2
     + y1  y2 - y1 y2  - y0 y1  + y0 y2 )/

    (x0 y1 - y0 x1 + x2 y0 - x2 y1 - y2 x0 + y2 x1)

                    2        2     2        2           2        2        2
yc = - 1/2 (- x0 x1  - x0 y1  + x0  x1 + y0  x1 - x2 x0  - x2 y0  + x2 x1

             2     2        2        2        2
      + x2 y1  + x2  x0 - x2  x1 + y2  x0 - y2  x1)/

     (x0 y1 - y0 x1 + x2 y0 - x2 y1 - y2 x0 + y2 x1)

>> Assuming the 3 points are (x0,y0), x1,y1), (x2, y2), the exact solution for 
>> the center (xc, yc) is:
[...]

	Quicky:  does .-1 imply that the following is a quick
	test for collinearity (i.e., compare the denominator
	from .-1 to zero)?

		(x0 y1 - y0 x1 + x2 y0 - x2 y1 - y2 x0 + y2 x1) == 0

	Dan

    That result could be written more perspicuously in terms of
    determinants...

    - Jim

>>    That result could be written more perspicuously in terms of
>>    determinants...

	I checked my dictionary to see if "perspicuously" or
	"perspicaciously" was correct ... you used the correct
	one there. :-)

	Dan