摘要：圆x2+y2-4x=0在点P(1.)处的切线方程为 A.x+y-2=0 B.x+y-4=0 C.x-y+4=0 D.x-y+2=0 解法一: x2+y2-4x=0 y=kx-k+ x2-4x+(kx-k+)2=0. 该二次方程应有两相等实根.即Δ=0.解得k=. ∴y-=(x-1).即x-y+2=0. 解法二:∵点(1.)在圆x2+y2-4x=0上. ∴点P为切点.从而圆心与P的连线应与切线垂直. 又∵圆心为(2.0).∴·k=-1. 解得k=.∴切线方程为x-y+2=0. 答案:D

网址：http://m.1010jiajiao.com/timu_id_2939745[举报]