Question

已知圆x²+y²-4x+4y+8-k=0关于直线x-y-2=0对称的圆是圆C,且圆C与直线3x+

已知圆x²+y²-4x+4y+8-k=0关于直线x-y-2=0对称的圆是圆C，且圆C与直线3x+4y-40=0相切，求实数k的值.

Answer 1

x²+y²-4x+4y+8-k=0
(x - 2)² + (y + 2)² = k
圆心A(2, -2)
AC与直线x-y-2=0垂直, x-y-2=0斜率为1, AC斜率为-1
AC的方程: y + 2 = -(x - 2), x + y = 0
AC与直线x-y-2=0的交点为B(1, -1)
B为AC的中点, 设C的坐标为C(a, b)
1 = (a + 2)/2, -1 = (b -2)/2
a = 0, b = 0
C(0, 0)为原点
圆C与直线3x+4y-40=0相切, C与该直线的距离为半径 r = |3*0 + 4*0 -40|/√(3^2 + 4^2) = 40/5=8
两圆半径相同, k = r^2 = 64