matlab程序实现三点确定一个圆（圆心，半径）

clc;clear all

A=input('请输入第一个坐标点的[x1,y1]：');

B=input('请输入第二个坐标点的[x2,y2]：');

C=input('请输入第三个坐标点的[x3,y3]：');

plot([A(1) B(1)],[A(2) B(2)],'b','linewidth',2);hold on

plot([A(1) C(1)],[A(2) C(2)],'b','linewidth',2);

plot([C(1) B(1)],[C(2) B(2)],'b','linewidth',2);

judge1=(B(1)-C(1))(B(2)-A(2))-(B(1)-A(1))(B(2)-C(2));

judge2=B(2)-A(2);

I1=0.5(C(2)-A(2))(B(2)-C(2))*(B(2)-A(2));

I2=0.5(A(1)^2-B(1)2)(B(2)-C(2));

I3=0.5(B(1)^2-C(1)2)(B(2)-A(2));

if(judge1==0)

fprintf('输入的三点构成一条直线，不能画一个圆。');

else

x0=(I1+I2+[I3](https://www.baidu.com/s?wd=I3&tn=44039180_cpr&fenlei=mv6quAkxTZn0IZRqIHckPjm4nH00T1dWnhNbuHmvnH-bm1bzmWnL0ZwV5Hcvrjm3rH6sPfKWUMw85HfYnjn4nH6sgvPsT6KdThsqpZwYTjCEQLGCpyw9Uz4Bmy-bIi4WUvYETgN-TLwGUv3EPWDLP1fYn1fvPj6zPjTsPHTz))/judge1;

if(judge2==0)

    y0=-(B(1)-C(1))*(x0-0.5*(B(1)+C(1)))/(B(2)-C(2))+0.5*(B(2)+C(2));

else

    y0=-(B(1)-A(1))*(x0-0.5*(B(1)+A(1)))/(B(2)-A(2))+0.5*(B(2)+A(2));

end

r=sqrt((x0-A(1))^2+(y0-A(2))^2);

theta=0:0.01:2*pi;

fprintf('\n圆的圆心是（%f,%f）\n',x0,y0);

fprintf('圆的半径是%f\n',r);

x=x0+r*cos(theta);

y=y0+r*sin(theta);

plot([x0,A(1)],[y0,A(2)],'g');

plot([x0,B(1)],[y0,B(2)],'g');

plot([x0,C(1)],[y0,C(2)],'g');

plot(x,y,'r','linewidth',2);

grid on

axis equal

end

运行后按照图片所示的格式输入坐标：

向左转

|

向右转

[图片上传失败...(image-303586-1511877287198)]

然后可以得到效果图哦！！！

向左转

|

向右转

[图片上传失败...(image-81957c-1511877287197)]