假设两点的坐标分别为A(x1,y1),B(x2,y2),圆的半径为rAB=√[(x1-x2)^2+(y1-y2)^2]圆心角为θ那么sin(θ/2)=(AB/2)/r=√[(x1-x2)^2+(y1-y2)^2]/(2r)所以θ=2arcsin(√[(x1-x2)^2+(y1-y2)^2]/(2r))弧长=θr=2r*arcsin(√[(x1-x2)^2+(y1-y2)^2]/(2r))
