python计算三个点构成的三角形的外切圆

推导细节请参考：https://blog.csdn.net/qq_17550379/article/details/78146201

import numpy as np


def circle(x1, y1, x2, y2, x3, y3):
    """
    :return:  x0 and y0 is center of a circle, r is radius of a circle
    """
    a = x1 - x2
    b = y1 - y2
    c = x1 - x3
    d = y1 - y3
    a1 = ((x1 * x1 - x2 * x2) + (y1 * y1 - y2 * y2)) / 2.0
    a2 = ((x1 * x1 - x3 * x3) + (y1 * y1 - y3 * y3)) / 2.0
    theta = b * c - a * d
    if abs(theta) < 1e-7:
        raise RuntimeError('There should be three different x & y !')
    x0 = (b * a2 - d * a1) / theta
    y0 = (c * a1 - a * a2) / theta
    r = np.sqrt(pow((x1 - x0), 2) + pow((y1 - y0), 2))
    return x0, y0, r


x0, y0, r = circle(48.955587, -122.745016, 0, 1, 4, 0)
print("圆心: " + str(x0) + " " + str(y0))
print("圆的半径: " + str(r))