圆与直线问题已知圆C:(x-1)^2+y^2=9,直线L:x-ky-3=0.问是否存在实数k,使得以直线L被圆C截得的弦

2个回答

  • 令k=1/a

    x-y/a-3=0

    y=-ax+3a

    代入

    (a^2+1)x^2-(2+6a^2)+(9a^2-8)=0

    x1+x2=(2+6a^2)/(a^2+1)

    y1+y2=-a(x1+x2)+6a=4a/(a^2+1)

    中点D[(1+3a^2)/(a^2+1),2a/(a^2+1)]

    OD^2=(9a^4+10a^2+1)/(a^2+1)^2

    x1x2=(9a^2-8)/(a^2+1)

    (x1-x2)^2=(x1+x2)^2-4x1x2=(28a^2+33)/(a^2+1)^2

    (y1-y2)^2=[(-ax1+3a)-(-ax2+3a)]^2=a^2(x1-x2)^2

    所以AB^2=(a^2+1)(x1-x2)^2=(28a^2+33)/(a^2+1)

    AB^2=4OD^2

    (28a^2+33)/(a^2+1)=4(9a^4+10a^2+1)/(a^2+1)^2

    a^2=29/8

    k^2=8/29

    x-√(8/29)y-3=0

    x+√(8/29)y-3=0