Question

已知直線l₁:mx+y-1-m=0(m∈R)和⊙C:x²+y²-2x=0.

(1)求直線l₁關(guān)于圓心C對稱的直線l₂的方程;

(2)證明:不論m為何實數(shù)時,直線l₂總與圓C有交點.

Answer 1

(1)解：由x²+y²-2x=0 (x-1)²+y²=1.

∴圓心C的坐標為(1,0).

設(shè)l₁上任一點P(x₁,y₁)關(guān)于點C(1,0)的對稱點為Q(x,y),

則

∴                                                               ①

又點P在直線l上,

∴mx₁+y₁-1-m=0.                                                          ②

①代入②得m(2-x)-y-1-m=0.

∴直線l₂的方程為mx+y+1-m=0.

(2)證明：由直線l₂的方程得m(x-1)+y+1=0,

∴x=1,y=-1時,不論m為何實數(shù)上式均成立.

∴l(xiāng)₂過定點A(1,-1).

又A在圓C上,故l₂總與圓C有交點.