如果实数x,y满足x^2+y^2-4x-5=0,求 （1） （y+6）/（x-5）的最大值 （2

x² + y² - 4x - 5 = 0
(x - 2)² + y² = 9
此为以C(2,0)为圆心,半径为3的圆.
(1)
令点A(5,-6),P(x,y),(y + 6)/(x - 5)的几何意义是AP的斜率；圆最右侧的点为Q(5,0),A,Q的横坐标相等,此时斜率不存在,其它情形下,斜率均小于0,(y + 6)/(x - 5)的最大值为过A的另一条切线的斜率,设为k,方程为y + 6 = k(x - 5)
kx - y - 5k - 6 = 0
C与其距离等于半径3 = |2k - 0 - 5k - 6|√(k² + 1)
k = -3/4
(2)
令t = y - x,y = x + t
(x - 2)² + y² = 9,(x - 2)² + (x + t)² = 9
(x + t)² = 5 + 4x - x²
t = -x + √(5 + 4x - x²) (舍去负号)
t' = -1 + (2 - x)/√(5 + 4x - x²) = 0
x = 2 - (3√2)/2时,y - x取最大值3√2 - 2
(3)
x² + y² - 4x - 5 = 0
x² + y² = 4x + 5,2 - 3 ≤ x ≤ 2 + 3,-1 ≤ x ≤ 5
x = 5时,x² + y²取最大值25