cutter@wpi.wpi.edu (Jeffrey T LeBlanc) (08/02/89)
Hello
Does anyone out there have an algorithm handy that, when given the
coordinates of three XY points can return the circle that would fall on
them? Any help along those lines would be appreciated.
CutterNU113738@NDSUVM1.BITNET (08/04/89)
If you're looking for a Circle algorithm and you have FTP capability on your machine, you might want to anonymous FTP albanycs.albany.edu (128.204.1.4). In the /pub/graphics directory there is large collection of old comp.graphics articles and code that were posted at one time or another here. There's alot to look at for old postings. Jeff Bakke NU113738@NDSUVM1.BITNET
myers@hpldola.HP.COM (Dan Myers) (08/05/89)
Jeffrey T LeBlanc writes: > Does anyone out there have an algorithm handy that, when given the >coordinates of three XY points can return the circle that would fall on >them? Any help along those lines would be appreciated. This is pretty off-the-cuff, but is based on the fact that the distance from any point on the perpendicular bisector of a line segment is equidistant from the original segments endpoints: Given: 3 points A, B, and C (in XY coords) Algorithm: If any of A,B,or C are duplicates, return the circle centered on the midpoint of the segment joining the two distinct endpoints with radius equal to 1/2 their distance. Determine the equations for the perpendicular bisectors of any TWO pairs of points( say AB and BC ). Call them perpAB and perpBC. If slope(perpAB) equals slope( perpBC ), return an "infinite" circle (i.e., the line passing through all points since they are collinear) Find the intersection of perpAB and perpBC (guaranteed exist since we made it past the last step...). Call it centerpoint. Return the circle with center centerpoint and radius equal to the distance from centerpoint to ANY of the given points (e.g. distance(centerpoint, A) ). Details are left as an exercise to the reader :^) Hope this helps... Dan Myers hplabs.hp.com!hpldola!myers
daveb@pogo.WV.TEK.COM (Dave Butler) (08/15/89)
Just saw the discussion about circles: Dan Myers writes: > Jeffrey T LeBlanc writes: > >> Does anyone out there have an algorithm handy that, when given the >>coordinates of three XY points can return the circle that would fall on >>them? Any help along those lines would be appreciated. Any two points on a circle form a cord of that circle. A line that is perpendicular to a cord and bisects that cord also bisects the circle (and therefore passes through the center of the circle). Two unique cords, will have two unique perpendicular bisection lines, both of which pass through the center of the circle. Therefore calculate the formula for these lines and then calculate their intersection point, because that's where the center of the circle is located. Later, Dave Butler Why does this magnificent applied science, which saves work and makes life easier, bring us so little happiness? The simple answer runs: Because we have not yet learned to make sensible use of it. Albert Einstein