Find Circle Center and RadiusDate: 09/21/2001 at 08:51:17 From: Alex Subject: Geometry If you are given three sets of (x,y) coordinates that lie on the circumference of a circle, how do you figure out the center and radius of the circle? Date: 09/21/2001 at 13:43:40 From: Doctor Rob Subject: Re: Geometry Thanks for writing to Ask Dr. Math, Alex. You can start with the three-point form of the equation of the circle, which can be found on the following page from our Frequently Asked Questions (FAQ): http://mathforum.org/dr.math/faq/formulas/faq.analygeom_2.html#twocircles | x^2+y^2 x y 1| |x1^2+y1^2 x1 y1 1| |x2^2+y2^2 x2 y2 1| = 0. |x3^2+y3^2 x3 y3 1| Here (x1,y1), (x2,y2), and (x3,y3) are the three given points. Expand this 4-by-4 determinant, complete the squares on x and y, and transform it into the center-radius 1form (x-h)^2 + (y-k)^2 = r^2. Then the center is (h,k) and the radius is r. Explicit expressions for h, k, and r^2 are as follows: |x1^2+y1^2 y1 1| |x2^2+y2^2 y2 1| |x3^2+y3^2 y3 1| h = ----------------, |x1 y1 1| 2*|x2 y2 1| |x3 y3 1| |x1 x1^2+y1^2 1| |x2 x2^2+y2^2 1| |x3 x3^2+y3^2 1| k = ----------------. |x1 y1 1| 2*|x2 y2 1| |x3 y3 1| |x1 y1 x1^2+y1^2| |x2 y2 x2^2+y2^2| |x3 y3 x3^2+y3^2| r^2 = h^2 + k^2 + -----------------. |x1 y1 1| |x2 y2 1| |x3 y3 1| In expanded form, these are: (x1^2+y1^2)(y2-y3) + (x2^2+y2^2)(y3-y1) + (x3^2+y3^2)(y1-y2) h = ------------------------------------------------------------ 2(x1y2 - x2y1 - x1y3 + x3y1 + x2y3 - x3y2) (x1^2+y1^2)(x3-x2) + (x2^2+y2^2)(x1-x3) + (x3^2+y3^2)(x2-x1) k = ------------------------------------------------------------ 2(x1y2 - x2y1 - x1y3 + x3y1 + x2y3 - x3y2) (x1^2+y1^2)(x2y3-x3y2) + (x2^2+y2^2)(x3y1-x1y3) + (x3^2+y3^2)(x1y2-x2y1) r^2 = h^2 + k^2 + --------------------------------------- x1y2 - x2y1 - x1y3 + x3y1 + x2y3 - x3y2 - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ Date: 05/17/2002 at 10:31:05 From: Michael Greene Subject: Equation for a circle given (x,y) of 3 points I have posed this question to geometry and Algebra 2 classes for the past 3-4 years, and the geometric solution may be simpler for many students who recall the geometry and may not be as comfortable with systems of equations: Consider A(-2,1) B(2,3) C(0,-5) a) The center of a circle may be found by the intersection of the perpendicular bisectors of 2 chords. b) Find the slopes and midpoints of 2 chords: AB: (0,2) m = 1/2 BC: (1,-1) m = 4 c) Find the slopes of the perpendicular lines. d) Find the equations of the perpendicular bisectors: Through AB, y = -2x+2 Through BC, y = (-1/4)x -3/4 e) Find the intersection of those lines -2x +2 = (-1/4)x -3/4 -8x +8 = -x -3 -7x = -11 x = 11/7 and y = -8/7 This isn't so much a question as an alternate solution, but I wanted to give something back to the Math Forum after referring some of my students to the archives and finding new ideas for our classes. Thanks, Mike Greene |
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