已知圆C:x^2+y^2+Dx-6y+1=0上有两点P,Q关于直线x-y+4=0对称

1个回答

  • 1.圆方程:(x + D/2)² + (y - 3)² = 8 + D²/4

    有两点P,Q关于直线x-y+4=0对称,圆心(-D/2,3)在直线上:-D/2 - 3 + 4 = 0,D = 2

    r² = 8 + 4/4 = 9

    r = 3

    2.OP垂直于OQ,显然,PQ为一个直径,且与x-y+4=0 (斜率1)垂直,PQ斜率= -1

    圆心(-1,3)

    直线PQ方程:y - 3 = -1(x + 1),y = 2 - x

    3.弦长最短,为0,圆心与直线的距离为半径:

    r = 3 = |-2m + 1 - 3m + 3 + 8m - 6}/√[(2m - 1)² + (m - 1)²] = |3m -2|/√[(2m - 1)² + (m - 1)²]

    36m² - 51m + 14 = 0

    (12m - 7)(3m - 2) = 0

    m = 7/12或m = 2/3