已知圆C:x^2+y^2-6mx-2(m-1)y+10m^2-2m-24=0,(1)求证无论m为何值,圆心在一条直线L上

(1)
x²+y²-6mx-2(m-1)y+10m²-2m-24=0
(x - 3m)² + [y -(m-1)]² = 25
圆心C(3m,m-1),半径5
令C(x,y):x = 3m,y = m - 1
消去m:y = x/3 - 1
圆心C在斜率1/3为的直线上
(2)
令直线为y = x/3 + c,x -3y + 3c = 0
C与该直线的距离为d = |3m -3(m-1) + 3c|/√(1² + 3²) = 3|c+1|/√10
直线被圆所截得的线段长l = 2*√(r² - d²)
r,d均与m无关,l也与m无关,