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：
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然后可以得到效果图哦！！！
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