[从头学数学] 第173节 圆与方程

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。

<span style="font-size:18px;">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*25
78.5</span>

知道三个坐标求外接圆的方程，是这样求的：

<span style="font-size:18px;">	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));
}
</span>

再来试一个：

<span style="font-size:18px;">        //已知三角形顶点阵列  
        var array = [[-5, -2], [5, -2], [0, 5]]; </span>

这个还是比較好玩的。

圆的一般方程：

<span style="font-size:18px;">//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)]];
}</span>

知道了能够做些什么呢？

假设如今已经知道了D、E、F的值：

<span style="font-size:18px;">	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);
	}</span>

这就是过那三个点的圆。

D、E、F能够这样来求：

<span style="font-size:18px;">	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+'<br/>');  
			matrix.print(ma);  
		document.write('da = ' + da+'<br/>');  
			matrix.print(mb);  
		document.write('db = '+db+'<br/>');  
		  	matrix.print(mc);  
		document.write('dc = '+dc+'<br/>');  
		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+'<br/>');     
	
	
	}
	</span>

小伟把求DEF的过程炼制成了工具，来玩玩吧：

<span style="font-size:18px;">//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+'<br/>');  
			matrix.print(ma);  
		document.write('da = ' + da+'<br/>');  
			matrix.print(mb);  
		document.write('db = '+db+'<br/>');  
		  	matrix.print(mc);  
		document.write('dc = '+dc+'<br/>');  
		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+'<br/>');   */  
}
		</span>

<span style="font-size:18px;">	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);
	}
	</span>

过A。 B， C三个点的圆，没错吧。

再玩一局：

<span style="font-size:18px;">var array = [[-5, -2], [-7, -8], [4, 5]]; </span>

这三个点是偏安一方啊。

直线和圆的位置关系：

<span style="font-size:18px;">	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);
	}</span>

这两个圆的位置关系：

<span style="font-size:18px;">	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);
	}</span>

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。

<span style="font-size:18px;">//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+'<br/>');  
			matrix.print(ma);  
		document.write('da = ' + da+'<br/>');  
			matrix.print(mb);  
		document.write('db = '+db+'<br/>');  
		  	matrix.print(mc);  
		document.write('dc = '+dc+'<br/>');  
		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+'<br/>');   */  
}
	</span>

本节到此结束。欲知后事怎样。请看下回分解。