用python实现 线段与线段相交于点，一个圆心位于点的圆与和相切。已知点的坐标和圆半径，需要编写程序计算圆与线段和线段的两个切点的坐标，弦长以及外距

您好，我可以回答这个问题。可以使用 Python 中的数学库来计算圆与线段和线段的两个切点的坐标，弦长以及外距。具体实现方法可以参考以下代码：

import math

# 线段与线段相交于点的函数
def intersection(p1, p2, p3, p4):
    x1, y1 = p1
    x2, y2 = p2
    x3, y3 = p3
    x4, y4 = p4
    d = (x1-x2)*(y3-y4) - (y1-y2)*(x3-x4)
    if d == :
        return None
    x = ((x1*y2-y1*x2)*(x3-x4) - (x1-x2)*(x3*y4-y3*x4)) / d
    y = ((x1*y2-y1*x2)*(y3-y4) - (y1-y2)*(x3*y4-y3*x4)) / d
    if x < min(x1, x2) or x > max(x1, x2) or x < min(x3, x4) or x > max(x3, x4):
        return None
    if y < min(y1, y2) or y > max(y1, y2) or y < min(y3, y4) or y > max(y3, y4):
        return None
    return (x, y)

# 计算圆与线段和线段的两个切点的坐标，弦长以及外距的函数
def circle_line_intersection(p1, p2, center, radius):
    x1, y1 = p1
    x2, y2 = p2
    xc, yc = center
    dx = x2 - x1
    dy = y2 - y1
    a = dx**2 + dy**2
    b = 2*dx*(x1-xc) + 2*dy*(y1-yc)
    c = (x1-xc)**2 + (y1-yc)**2 - radius**2
    delta = b**2 - 4*a*c
    if delta < :
        return None
    t1 = (-b + math.sqrt(delta)) / (2*a)
    t2 = (-b - math.sqrt(delta)) / (2*a)
    if t1 <  or t1 > 1:
        t1 = None
    if t2 <  or t2 > 1:
        t2 = None
    p1 = None
    p2 = None
    if t1 is not None:
        x = x1 + t1*dx
        y = y1 + t1*dy
        p1 = (x, y)
    if t2 is not None:
        x = x1 + t2*dx
        y = y1 + t2*dy
        p2 = (x, y)
    if p1 is None and p2 is None:
        return None
    elif p1 is None:
        return p2
    elif p2 is None:
        return p1
    else:
        d1 = math.sqrt((p1[]-xc)**2 + (p1[1]-yc)**2)
        d2 = math.sqrt((p2[]-xc)**2 + (p2[1]-yc)**2)
        chord = 2*math.sqrt(radius**2 - ((d1+d2)/2)**2)
        distance = abs((p2[]-p1[])*(p1[1]-yc) - (p1[]-xc)*(p2[1]-p1[1])) / math.sqrt((p2[]-p1[])**2 + (p2[1]-p1[1])**2)
        return p1, p2, chord, distance

# 示例
p1 = (, )
p2 = (1, 1)
center = (.5, .5)
radius = .5
result = circle_line_intersection(p1, p2, center, radius)
print(result)

希望能对您有所帮助。