给定一个圆上的2个点及其之间的角度，如何找到中心（在python中）？

Question

我在圆上有2个点，并且它们之间有一个角度，我想找到这样定义的圆的中心（嗯，最好是两个中心）。

def find_center(p1,p2,angle):
  # magic happens... What to do here?
  return (center_x, center_y)

Answer 1

这是我的测试代码解决方案

from pylab import *
from numpy import *

def find_center(p1, p2, angle):
    # End points of the chord
    x1, y1 = p1 
    x2, y2 = p2 

    # Slope of the line through the chord
    slope = (y1-y2)/(x1-x2)

    # Slope of a line perpendicular to the chord
    new_slope = -1/slope

    # Point on the line perpendicular to the chord
    # Note that this line also passes through the center of the circle
    xm, ym = (x1+x2)/2, (y1+y2)/2

    # Distance between p1 and p2
    d_chord = sqrt((x1-x2)**2 + (y1-y2)**2)

    # Distance between xm, ym and center of the circle (xc, yc)
    d_perp = d_chord/(2*tan(angle))

    # Equation of line perpendicular to the chord: y-ym = new_slope(x-xm)
    # Distance between xm,ym and xc, yc: (yc-ym)^2 + (xc-xm)^2 = d_perp^2
    # Substituting from 1st to 2nd equation for y,
    #   we get: (new_slope^2+1)(xc-xm)^2 = d^2

    # Solve for xc:
    xc = (d_perp)/sqrt(new_slope**2+1) + xm

    # Solve for yc:
    yc = (new_slope)*(xc-xm) + ym

    return xc, yc

if __name__=='__main__':
    p1 = [1., 2.]
    p2 = [-3, 4.]
    angle = pi/6
    xc, yc = find_center(p1, p2,angle)

    # Calculate the radius and draw a circle
    r = sqrt((xc-p1[0])**2 + (yc-p1[1])**2)
    cir = Circle((xc,yc), radius=r,  fc='y')
    gca().add_patch(cir)

    # mark p1 and p2 and the center of the circle
    plot(p1[0], p1[1], 'ro')
    plot(p2[0], p2[1], 'ro')
    plot(xc, yc, 'go')

    show()