center of the circle.

[martin]

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Dim $ax = 48
Dim $ay = 396

Dim $bx = 275
Dim $by = 497

Dim $cx = 486
Dim $cy = 326
radar_center( $ax, $ay, $bx, $by, $cx, $cy )
Func radar_center( $ax, $ay, $bx, $by, $cx, $cy )
    

    
    ;slopes
        
        $m1 = ( ( $by - $ay ) / ( $bx - $ax ) );
        $m2 = ( ( $cy - $by ) / ( $cx - $bx ) )
        $m1 = -$m1
        $m2 = -$m2
        
    ;midpoints
        
        $ab_mid_x = ( $bx + $ax ) / 2
        $ab_mid_y = ( $by + $ay ) / 2
        
        $bc_mid_x = ( $cx + $bx ) / 2
        $bc_mid_y = ( $cy + $by ) / 2
    
    ;c = y - mx ;     y = mx + c
    
        $c1 = $ab_mid_y - $m1 * $ab_mid_x
        $c2 = $bc_mid_y - $m2 * $bc_mid_x
        
    ;center of the circle
    ;shift signs for perpendicular functions.

    
        $center_x = ( $c1 - $c2 ) / ( $m2 - $m1 )
        $center_y = $m2 * $center_x + $c2
        
        
        MsgBox ( 0 , "title", $center_x )
        MsgBox ( 0 , "title", $center_y )
    
    ;Return $center_x & $center_y
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Points to watch for

You need to add a check in your calc for horizontal or vertical lines so that you don't try to divide by zero.

In these cases you just have to say X= K or Y = K where K is some constant.

You have to check that the 2 lines you are using are not parallel. If they are then use the end of one to the end of the other as the second line, assuming they are not the same line of course.

Edit:removed colur tags which don't work inside autoit tags.

Edited May 13, 2007 by martin

[alrick]

We can get the three red points by default, and we draw two lines, from point 1 to point 2, and from point 2 and point 3.

from the center of those two lines draw two lines perpenticular to the the two lines.

and then at the intersection of these two new lines, this is the center of the circle.

ok, so we get coordinates of 3 (x,y) points, how do we program au3 to find the center of the circle?

If the lines that cut the circle are equal in length then the solution is easy.

For example;-

Line 1 cuts the circle at 100,50 & 500,50. Line 2 cuts the circle at 500,50 & 500,450 - both lines are 400 pixels long at 90 degrees to each other.

So the centre is x1 + (x2 - x1)/2 , y2 + (y2 - y1)/2 = 300,250

Edited May 13, 2007 by alrick

[martin]

If the lines that cut the circle are equal in length then the solution is easy.

For example;-

Line 1 cuts the circle at 100,50 & 500,50. Line 2 cuts the circle at 500,50 & 500,450 - both lines are 400 pixels long at 90 degrees to each other.

So the centre is x1 + (x2 - x1)/2 , y2 + (y2 - y1)/2 = 300,250

Your example is based on the lines being horizontal and vertical, not whether they are equal lengths.

If line1 is vertical then the centre y coord of the circle is at the mid point of line1, and if a line2 is horizontal then the centre x coord of the circle is the mid point of line2. It doesn't matter whether they are of equal lengths.

Say line1 is x1,y1 to x2,y1

line2 is x3,y2,x3,y3

Then you simply have the centre is (x1 + x2)/2, (y2 + y3) /2

Not

x1 + (x2 - x1)/2 , y2 + (y2 - y1)/2 which does not give 300,250 - you meant y2 +(y1 - y2)/2

But the whole point is in finding a general solution when the lines are anything including vertical and horizontal.

Edited May 13, 2007 by martin

[qwertylol]

I guess it can be done this way?

if two points share the same x or y, then it's horizontal or vertical ?

[martin]

I guess it can be done this way?

if two points share the same x or y, then it's horizontal or vertical ?

Don't understand what you mean there.

Did you see my earlier post where I got your radar_center function to work?

[alrick]

Your example is based on the lines being horizontal and vertical, not whether they are equal lengths.

If line1 is vertical then the centre y coord of the circle is at the mid point of line1, and if a line2 is horizontal then the centre x coord of the circle is the mid point of line2. It doesn't matter whether they are of equal lengths.

Say line1 is x1,y1 to x2,y1

line2 is x3,y2,x3,y3

Then you simply have the centre is (x1 + x2)/2, (y2 + y3) /2

Not

x1 + (x2 - x1)/2 , y2 + (y2 - y1)/2 which does not give 300,250 - you meant y2 +(y1 - y2)/2

But the whole point is in finding a general solution when the lines are anything including vertical and horizontal.

Yes, I was too quick to post before thinking about it.

The solution is 2 isosceles triangles that have common length sides, or two pairs of 90 degree triangles, IYSWIM

[martin]

Yes, I was too quick to post before thinking about it.

The solution is 2 isosceles triangles that have common length sides, or two pairs of 90 degree triangles, IYSWIM

Yes.

I didn't notice before but that post of yours was your first, so welcome to the forum!