如何计算三维空间中圆与直线的最短距离？

import numpy as np

# 已知条件
o = np.array([0.3501, -0.0881, -4.8466])  # 圆心
n = np.array([0.4163, -0.8326, -0.3653])  # 圆的法向量
r = 1.34954  # 圆的半径
a = np.array([3.1932, -0.9005, 0.8082])  # 直线上的点A
b = np.array([1.9885, -0.9691, -0.8353])  # 直线上的点B

# 计算直线方向向量
ab = b - a
ab_direction = ab / np.linalg.norm(ab)

# 计算直线法向量
normal_to_line = np.cross(n, ab_direction)
normal_to_line = normal_to_line / np.linalg.norm(normal_to_line)

# 计算圆心O在法平面上的投影
o_projected = o - np.dot(o - a, normal_to_line) * normal_to_line

# 计算投影点到直线的距离
t = np.dot(o_projected - a, ab_direction) / np.linalg.norm(ab_direction)**2
closest_point_on_line = a + t * ab_direction
shortest_distance = np.linalg.norm(o_projected - closest_point_on_line)

# 计算点P坐标 (近似解，因为点P可能不在O'与直线的连线上)
p_on_circle = o + (closest_point_on_line - o) * (r / np.linalg.norm(closest_point_on_line - o))


print("圆上离直线最近的点P的坐标:", p_on_circle)
print("圆与直线的最短距离:", shortest_distance)