如图,在直角坐标系xoy中,△ABC的顶点坐标为A(—8,0),B(3,0),C(0,4),动点P从点A出发,沿着AB以

如图,在直角坐标系xoy中,△ABC的顶点坐标为A(—8,0),B(3,0),C(0,4),动点P从点A出发,沿着AB以每秒1个单
沿着AB以每秒1个单位长度的速度向终点B运动；动点Q从点B出发,沿着射线BC,以每秒1个单位长度的速度运动,当点P到达B时,点Q也停止运动,P,Q两点同时开始运动,设运动时间为t秒.
（1）当PQ⊥x轴,求此时t的值；
（2）设△BPQ的面积为S,求S关于t的函数关系式；
（3）当△APQ为等腰三角形时,求t的值；
（4）设△APQ的外接圆的圆心为M,当点C在圆M外时,写出t的取值范围.

(1)
AB = 11,0 ≤ t ≤ 11
t秒时,P(-8 + t,0),BQ = t
OB = 3,OC = 4,BC = 5
cos∠ABC = OB/BC = 3/5; sin∠ABC = OC/BC = 4/5
Q的横坐标 = OB - BQcos∠ABC = 3 - 3t/5
Q的纵坐标 = BQsin∠ABC = 4t/5
当PQ⊥x轴时,P,Q的横坐标相等:-8 + t = 3 - 3t/5
t = 55/8
(2)
S = (1/2)PB*Q的纵坐标
= (1/2)*(3 + 8 - t)(4t/5)
= (2t/5)(11 - t)
= -2t²/5 + 22t/5 (0 ≤ t ≤ 11)
(3)
AP² = t²
AQ² = (3 - 3t/5 + 8)² + (4t/5 - 0)² = (11 - 3t/5)² + (4t/5)²
PQ² = (-8 + t - 3 + 3t/5)² + (4t/5)² = (8t/5 - 11)² + (4t/5)²
(i)A为顶点,AP² = AQ²
t² = (11 - 3t/5)² + (4t/5)²
t = 55/6 < 11,符合要求
(ii) P为顶点,AP² = PQ²
t² = (8t/5 - 11)² + (4t/5)²
t² - 16t + 55 = (t - 5)(t - 11) = 0
t = 5或t = 11,均符合要求(0 ≤ t ≤ 11)
(iii)Q为顶点,AQ² = PQ²
t² - 10t = 0
t = 10 (舍去t = 0)
(4)
AP的中点为U(-8 + t/2,0),AP的中垂线为l1:x = -8 + t/2 (i)
PQ的中点为V((2t - 25)/10,2t/5)
PQ的斜率k = (4t/5 - 0)/(3 - 3t/5 + 8 - t) = 4t/(55 - 8t)
PQ的中垂线斜率k' = -1/k = (8t - 55)/(4t)
PQ的中垂线为l2:y - 2t/5 = [(8t - 55)/(4t)][x - (2t - 25)/10] (ii)
从(i)(ii)可得外接圆的圆心M(-8 + t/2,m),这里m = (8t² - 121t + 605)/(8t)
外接圆半径r
r² = MA² = (-8 + t/2 + 8)² + (m - 0)² = t²/4 + m²
CM² = (-8 + t/2 - 0)² + (m - 4)² = t²/4 + m² + 64 - 8t - 8m + 16
= r² + 80 - 8(t + m)
点C在圆M外时,CM² - r² = 80 - 8(t + m) > 0
t + m < 10
t + (8t² - 121t + 605)/(8t) < 10
16t² - 201t + 605 < 0
(16t - 121)(t - 5) < 0
5 < t < 121/16
没有复查(4),请自己验证.
再问： 第3题的3是如何求出的，第4题能用初中知识吗
再答： AQ² = (3 - 3t/5 + 8)² + (4t/5 - 0)² = (11 - 3t/5)² + (4t/5)² PQ² = (-8 + t - 3 + 3t/5)² + (4t/5)² = (8t/5 - 11)² + (4t/5) (iii)Q为顶点, AQ² = PQ² (11 - 3t/5)² + (4t/5)² = (8t/5 - 11)² + (4t/5)² (11 - 3t/5)² = (8t/5 - 11)² 11 - 3t5 = 8t/5 - 11, 11t/5 = 22, t = 10 或 11 - 3t/5 = 11 - 8t/5, t = 0, 舍去