Question

已知点p（5,-3）点Q是圆x²+y²-2x-4y-3=0上的动点，点M为线段PQ的中点，若M到直

已知点p（5,-3）点Q是圆x²+y²-2x-4y-3=0上的动点，点M为线段PQ的中点，若M到直线x-y+a=0的最小值为5根号2/2，求a的值

Answer 1

x²+y²-2x-4y-3=0(x - 1)² + (y - 2)² = 8
令Q(u, v), M(x, y)
(u - 1)² + (v - 2)² = 8 (1)
x = (u + 5)/2, u = 2x - 5
y = (v - 3)/2, v = 2y + 3
代入(1)得到M的轨迹：(2x - 6)² + (2y +1)² = 8
(x - 3)² + (y + 1/2)² = 2
此为以A(3, -1/2)为圆心,半径为r = √2的圆
M到直线x-y+a=0的最小距离为5√2/2，则A与其最小距离为d = 5√2/2 + r = 7√2/2
d = |3 + 1/2 + a|/√2 = 7√2/2
|a + 7/2| = 7
a = 7/2或a = -21/2