已知圆C: x^2+y^2-2x-2y+1=0,直线L:y=kx,且L与圆C交与P、Q两点,点M(0,b)满足MP垂直M

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  • C:x^2+y^2-2x-2y+1=0 => (x-1)^2 + (y-1)^2 =1

    圆心 (1,1),半径 R=1.圆与y轴相切于(0,1),与x轴相切于(1,0).

    1.当b=1时,M(0,1),MP垂直MQ,=> PQ是圆的直径

    =》 PQ在直线 y=x 上,k=1

    2.k>3,y=kx 与圆的交点满足:(x-1)^2 + (kx-1)^1=1

    (k^2+1) x^2 - 2(1+k) x +1 = 0

    => x1,x2 = [1+k ±√(2k)] /(k^2+1),x1+x2 = 2(1+k)/(k^2+1),x1* x2 = 1/(k^2+1)

    MP垂直MQ => [ (b - k * x1) /(-x1)] * [ (b - k * x2) /(-x2)] = -1

    => (b - k * x1) * (b - k * x2) = - x1 * x2

    => b^2 - k(x1+x2) * b + (k^2+1) x1* x2 = 0

    => b^2 - 2k(1+k)/(k^2+1) * b + 1 =0

    => b1,b2 = .= [ k(k+1) ± √ ( 2k^3 - k^2 -1) ] / (k^2+1)