已知曲线C1的方程为x^2-y^2/8=1(x>=o,y>=0),圆C2的方程为（x-3）^2+y^2=1,

C2的圆心为C(3, 0), 半径为r = 1
x² - y²/8 = 1, x ≥ 0, y ≥ 0为双曲线在第一象限的部分.
|AC| = r = 1
|AB| = √3
AC⊥AB, |BC|² = |AC|² + |AB|² = 1 + 3 = 4
即B为圆(x - 3)² + y² = 4和C1的交点.
消去y得: (3x + 1)(x - 1) = 0
x = 1 (舍去x = -1/3)
B(1, 0)
切线: y - 0 = k(x - 1), kx - y - k = 0
C与切线的距离d = r = 1 = |3k - 0 - k|/√(k² + 1)
3k² = 1
k = √3/3 (舍去k = -√3/3 < 0)
请采纳.