直线与圆 (21 13:46:14)

4个回答

  • 1.圆的圆心为(-3,-3),由对称知圆C圆心为(0,0).由圆C过(1,1)点知圆C半径为根号2.圆C方程:x2+y2=2

    2.可设直线PA为 y-1=k(x-1)

    则直线PB为 y-1=-k(x-1)

    设A为(x1,y1),B为(x1,y2)

    联立直线PA和圆C,得 [k(x-1)+1]^2+x^2=2

    即 (k^2+1)*x^2-2k(k-1)x+k^2-2k-1=0

    由伟达定理得 x1*1=(k^2-2k-1)/(k^2+1),

    x1=(k^2-2k-1)/(k^2+1)

    y1=k(x1-1)+1=(-k^2-2k+1)/(k^2+1),

    同理可得x2=(k^2+2k-1)/(k^2+1) ,y2 =(-k^2+2k+1)/(k^2+1),

    AB斜率为(y1-y2)/(x1-x2)=1

    与OP斜率相等,所以AB//OP