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厄,大家都是数学地,我来个机械地In Reply To:给出公式! 不需要解析几何! Posted by:810974380 at 2009-08-03 15:46:30 已知圆的方程(x-a)^2+(y-b)^2=r^2 又知道三个不共线点,联立3元2次方程。 用数学软件求得:r= [ 1/2*((-2*x2*x3-2*y2*y3+y2^2+y3^2+x2^2+x3^2)*(x1^2-2*y1*y3+y1^2+y3^2+x3^2-2*x1*x3)*(x1^2-2*x2*x1+y2^2-2*y1*y2+x2^2+y1^2))^(1/2)/(x1*y2-x1*y3-y1*x2+y1*x3-x3*y2+y3*x2)] [ -1/2*((-2*x2*x3-2*y2*y3+y2^2+y3^2+x2^2+x3^2)*(x1^2-2*y1*y3+y1^2+y3^2+x3^2-2*x1*x3)*(x1^2-2*x2*x1+y2^2-2*y1*y2+x2^2+y1^2))^(1/2)/(x1*y2-x1*y3-y1*x2+y1*x3-x3*y2+y3*x2)] 这两个解等值反号,剩下的... 仔细一看,上式可以从abc/4s转换过来,天下数学本是一家。 Followed by: Post your reply here: |
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