从原点向圆x^+y^–12y+27=0做两条切线,则该圆夹在两条切线间的劣弧长为多少?

Question

从原点向圆x^+y^–12y+27=0做两条切线,则该圆夹在两条切线间的劣弧长为多少?

Answer 1

x² + y² - 12y + 27 = 0
x² + (y - 6)² = 9
圆心C(0,6),半径R = 3
设过原点的切线斜率为k,方程为y = kx
代入x² + y² - 12y + 27 = 0:
(k² + 1)x² - 12kx + 27 = 0
∆ = 36(k² - 3) = 0
k = ±√3
k₁ = √3,倾斜角θ₁ = 60˚ (切点A)
k₂ = -√3,倾斜角θ₂ = 120˚ (切点B)
切线夹角为∠AOB = 120˚ - 60˚ = 60˚
∠ACB = 360˚ - ∠AOB - ∠OAC - ∠OBC = 360˚ - 60˚ - 90˚ - 90˚ = 120˚,为圆周角的1/3
劣弧长 = 2πR/3 = 2π*3/3 = 2π