剧情提要：[机器小伟]在[project师阿伟]的陪同下进入了结丹初期的修炼。这次要修炼的目标是[圆与方程]。 
正剧開始：
星历2016年04月11日 15:56:41, 银河系厄尔斯星球中华帝国江南行省。 [project师阿伟]正在和[机器小伟]一起研究[圆与方程]。
已知圆心位置和半径来画圆，小伟用36边的多边形取代圆，
面积上略微小了一点。
if (1) { var r = 20; config.setSector(1,1,1,1); config.graphPaper2D(0, 0, r); config.axis2D(0, 0, 180); var transform = new Transform(); var a = 5, b = 5, r0 = 5; var array = shape.nEdge(a, b, r0, 36); var scale = r; //array = shape.angularSort(array); shape.areaDraw(transform.translate(array, -200/scale, -200/scale), 'red', scale); shape.strokeDraw([].concat(array), 'orange', scale); } >>> 3.14*2578.5
知道三个坐标求外接圆的方程，是这样求的：
if (1) {//求三角形的外心 var r = 20; var r0 = 5*r; config.setSector(1,1,1,1); config.graphPaper2D(0, 0, r); config.axis2D(0, 0, 180); var triangle = new Triangle(); var transform = new Transform(); //已知三角形顶点阵列 var array = [[5, 1], [7, -3], [2, -8]]; //进行缩放转换 // array = transform.scale(array, r); //三个顶点 var x1 = array[0][0], y1 = array[0][1], x2 = array[1][0], y2 = array[1][1], x3 = array[2][0], y3 = array[2][1]; //令 var A1 = 2*(x2-x1), B1 = 2*(y2-y1), C1 = x2*x2+y2*y2-x1*x1-y1*y1, A2 = 2*(x3-x2), B2 = 2*(y3-y2), C2 = x3*x3+y3*y3-x2*x2-y2*y2; //得到外心的坐标 var px = ((C1*B2)-(C2*B1))/((A1*B2)-(A2*B1)), py = ((A1*C2)-(A2*C1))/((A1*B2)-(A2*B1)); //外接圆半径 var a = distance2D(array[0], array[1]), b = distance2D(array[1], array[2]), c = distance2D(array[2], array[0]); var rOut = a*b*c/Math.sqrt(4*b*b*c*c-Math.pow((b*b+c*c-a*a), 2)); //document.write(px.toFixed(2)+', '+py.toFixed(2)+'; ' + rOut.toFixed(2)); var scale = r; shape.angleDraw([].concat(array), 'red', scale); var circle = shape.nEdge(px, py, rOut, 36); shape.strokeDraw([].concat(circle), 'blue', scale); var s = '外心: ['+px.toFixed(2)+' , ' + py.toFixed(2)+'] '; var s1 = '外接圆半径：'+rOut.toFixed(2); plot.setFillStyle('#FF2288'); plot.fillText(s, -270, -170, 300); plot.fillText(s1, -270, -140, 300); }//二维坐标中两点之间的距离function distance2D(pointA, pointB) { return Math.sqrt(Math.pow(pointA[0]-pointB[0], 2)+Math.pow(pointA[1]-pointB[1], 2));}
再来试一个：
//已知三角形顶点阵列 var array = [[-5, -2], [5, -2], [0, 5]]; 这个还是比較好玩的。
圆的一般方程：
//圆的一般方程function generalCircle(D, E, F) { //方程x^2+y^2+Dx+Ey+F=0; var rSquare = (D*D+E*E-4*F)/4; if (rSquare > 0) { return shape.nEdge(-D/2, -E/2, Math.sqrt(rSquare), 36); } else { return []; }}//直线的一般方程function generalLine(A, B, C) { //方程Ax+By+C = 0; return [[-100, (-100*A+C)/(-B)], [100, (100*A+C)/(-B)]];} 知道了能够做些什么呢？ 
假设如今已经知道了D、E、F的值：
if (1) { var r = 20; config.setSector(1,1,1,1); config.graphPaper2D(0, 0, r); config.axis2D(0, 0, 180); var scale = r; var circle = generalCircle(-8, 6, 0); shape.strokeDraw([].concat(circle), 'blue', scale); }
这就是过那三个点的圆。
D、E、F能够这样来求：
if (1) { //运用行列式解线性方程组 var matrix = new Matrix(); var matrixArray = new Array(); var rowArray = new Array(); var ma, mb, mc; //求过三点的圆的方程 //[x1, y1], [x2, y2], [x3, y3] // ==> 方程x^2+y^2+Dx+Ey+F=0; //解D, E, F var point = [[0,0],[1,1],[4,2]]; var x1 = point[0][0], y1 = point[0][1], x2 = point[1][0], y2 = point[1][1], x3 = point[2][0], y3 = point[2][1]; //三元一次方程组[[A1,B1,C1], [A2,B2,C2], ...] var a = [ [x1, y1, 1, x1*x1+y1*y1], [x2, y2, 1, x2*x2+y2*y2], [x3, y3, 1, x3*x3+y3*y3] ]; //三阶 var rank = 3; for (var i = 0; i < rank; i++) { matrixArray.push([a[i][0], a[i][1], a[i][2]]); } ma = matrix.deepCopy(matrixArray); for (var i = 0; i < rank; i++) { ma[i][0] = -a[i][3]; } mb = matrix.deepCopy(matrixArray); for (var i = 0; i < rank; i++) { mb[i][1] = -a[i][3]; } mc = matrix.deepCopy(matrixArray); for (var i = 0; i < rank; i++) { mc[i][2] = -a[i][3]; } var d, da, db, dc; d = matrix.delta(matrixArray); da = matrix.delta(ma); db = matrix.delta(mb); dc = matrix.delta(mc); matrix.print(matrixArray); document.write('d = ' + d+''); matrix.print(ma); document.write('da = ' + da+''); matrix.print(mb); document.write('db = '+db+''); matrix.print(mc); document.write('dc = '+dc+''); var s = 'D = da/d = '+ (da/d).toFixed(2)+', E = db/d = '+(db/d).toFixed(2) +', F = dc/d = '+(dc/d).toFixed(2); document.write(s+''); } 小伟把求DEF的过程炼制成了工具，来玩玩吧：
//求解圆的一般方程中的D、E、F三个系数//传入的是三个点的坐标阵列//传回[D, E, F];function solveDEF(point) { //运用行列式解线性方程组 var matrix = new Matrix(); var matrixArray = new Array(); var rowArray = new Array(); var ma, mb, mc; //求过三点的圆的方程 //[x1, y1], [x2, y2], [x3, y3] // ==> 方程x^2+y^2+Dx+Ey+F=0; //解D, E, F //var point = [[0,0],[1,1],[4,2]]; var x1 = point[0][0], y1 = point[0][1], x2 = point[1][0], y2 = point[1][1], x3 = point[2][0], y3 = point[2][1]; //三元一次方程组[[A1,B1,C1], [A2,B2,C2], ...] var a = [ [x1, y1, 1, x1*x1+y1*y1], [x2, y2, 1, x2*x2+y2*y2], [x3, y3, 1, x3*x3+y3*y3] ]; //三阶 var rank = 3; for (var i = 0; i < rank; i++) { matrixArray.push([a[i][0], a[i][1], a[i][2]]); } ma = matrix.deepCopy(matrixArray); for (var i = 0; i < rank; i++) { ma[i][0] = -a[i][3]; } mb = matrix.deepCopy(matrixArray); for (var i = 0; i < rank; i++) { mb[i][1] = -a[i][3]; } mc = matrix.deepCopy(matrixArray); for (var i = 0; i < rank; i++) { mc[i][2] = -a[i][3]; } var d, da, db, dc; d = matrix.delta(matrixArray); da = matrix.delta(ma); db = matrix.delta(mb); dc = matrix.delta(mc); return [da/d, db/d, dc/d]; /* matrix.print(matrixArray); document.write('d = ' + d+''); matrix.print(ma); document.write('da = ' + da+''); matrix.print(mb); document.write('db = '+db+''); matrix.print(mc); document.write('dc = '+dc+''); var s = 'D = da/d = '+ (da/d).toFixed(2)+', E = db/d = '+(db/d).toFixed(2) +', F = dc/d = '+(dc/d).toFixed(2); document.write(s+''); */ } if (1) { var r = 20; config.setSector(1,1,1,1); config.graphPaper2D(0, 0, r); config.axis2D(0, 0, 180); var array = [[-5, -2], [5, -2], [0, 5]]; var DEF = solveDEF(array); var scale = r; var circle = generalCircle(DEF[0], DEF[1], DEF[2]); shape.strokeDraw([].concat(circle), 'blue', scale); shape.angleDraw([].concat(array), 'red', scale); }
过A。 B， C三个点的圆，没错吧。
再玩一局：
var array = [[-5, -2], [-7, -8], [4, 5]];
这三个点是偏安一方啊。
直线和圆的位置关系：
if (1) { var r = 20; config.setSector(1,1,1,1); config.graphPaper2D(0, 0, r); config.axis2D(0, 0, 180); var scale = r; var circle = generalCircle(0, -2, -4); var line = generalLine(3, 1, -6); shape.strokeDraw([].concat(circle), 'blue', scale); shape.multiLineDraw([].concat(line), 'red', scale); }
这两个圆的位置关系：
if (1) { var r = 20; config.setSector(1,1,1,1); config.graphPaper2D(0, 0, r); config.axis2D(0, 0, 180); var scale = r; var circle_1 = generalCircle(2,8,-8), circle_2 = generalCircle(-4, -4,-2); shape.strokeDraw([].concat(circle_1), 'blue', scale); shape.strokeDraw([].concat(circle_2), 'red', scale); } 至于交点坐标倒底是什么，就要另外去求了。 图上看好像是[-1,1] 和[3, -1]，也不知对不正确。
[人叫板老师]也没有给出交点坐标。
这里(1.75-1.25)/(1.75+1.25) = 1/6。而1.75+1.25又恰好是PQ距离的一半。
算了，这样的规律没什么意思，直接化简得了。
//二维坐标中两点之间的距离function distance2D(pointA, pointB) { return Math.sqrt(Math.pow(pointA[0]-pointB[0], 2)+Math.pow(pointA[1]-pointB[1], 2));}//圆的一般方程function generalCircle(D, E, F) { //方程x^2+y^2+Dx+Ey+F=0; var rSquare = (D*D+E*E-4*F)/4; if (rSquare > 0) { return shape.nEdge(-D/2, -E/2, Math.sqrt(rSquare), 36); } else { return []; }}//直线的一般方程function generalLine(A, B, C) { //方程Ax+By+C = 0; return [[-100, (-100*A+C)/(-B)], [100, (100*A+C)/(-B)]];}//求解圆的一般方程中的D、E、F三个系数//传入的是三个点的坐标阵列//传回[D, E, F];function solveDEF(point) { //运用行列式解线性方程组 var matrix = new Matrix(); var matrixArray = new Array(); var rowArray = new Array(); var ma, mb, mc; //求过三点的圆的方程 //[x1, y1], [x2, y2], [x3, y3] // ==> 方程x^2+y^2+Dx+Ey+F=0; //解D, E, F //var point = [[0,0],[1,1],[4,2]]; var x1 = point[0][0], y1 = point[0][1], x2 = point[1][0], y2 = point[1][1], x3 = point[2][0], y3 = point[2][1]; //三元一次方程组[[A1,B1,C1], [A2,B2,C2], ...] var a = [ [x1, y1, 1, x1*x1+y1*y1], [x2, y2, 1, x2*x2+y2*y2], [x3, y3, 1, x3*x3+y3*y3] ]; //三阶 var rank = 3; for (var i = 0; i < rank; i++) { matrixArray.push([a[i][0], a[i][1], a[i][2]]); } ma = matrix.deepCopy(matrixArray); for (var i = 0; i < rank; i++) { ma[i][0] = -a[i][3]; } mb = matrix.deepCopy(matrixArray); for (var i = 0; i < rank; i++) { mb[i][1] = -a[i][3]; } mc = matrix.deepCopy(matrixArray); for (var i = 0; i < rank; i++) { mc[i][2] = -a[i][3]; } var d, da, db, dc; d = matrix.delta(matrixArray); da = matrix.delta(ma); db = matrix.delta(mb); dc = matrix.delta(mc); return [da/d, db/d, dc/d]; /* matrix.print(matrixArray); document.write('d = ' + d+''); matrix.print(ma); document.write('da = ' + da+''); matrix.print(mb); document.write('db = '+db+''); matrix.print(mc); document.write('dc = '+dc+''); var s = 'D = da/d = '+ (da/d).toFixed(2)+', E = db/d = '+(db/d).toFixed(2) +', F = dc/d = '+(dc/d).toFixed(2); document.write(s+''); */ } 本节到此结束。欲知后事怎样。请看下回分解。