已知线段AB的长为3,动点P满足|PA|=2|PB|.

(1) 取A为原点,B在+x轴上,A(0,0),B(3,0)
P(x,y)
√[(x - 0)² + (y - 0)²] = 2√[(x - 3)² + (y - 0)²]
平方并整理得:(x - 4)² + y² = 4
r = 2,圆心D(4,0)
(2)
设直线的斜率为k (圆关于x轴对称,不妨设k > 0)
直线方程y = kx,kx - y = 0
B与直线的距离为h = |3k|/√(k² + 1) = 3k/√(k² + 1)
D与直线的距离为d = |4k|/√(k² + 1) = 4k/√(k² + 1)
MN = 2√(r² - d²) = 2√[4 - 16k²/(k² + 1)] = 4√[(1 - 3k²)/(1 + k²)]
S = (1/2)MN*h
= (1/2)*[3k/√(k² + 1)]*4√[(1 - 3k²)/(1 + k²)]
= 6k[√(1 - 3k²)]/(1 + k²)
对k求导得:S' = 6(1 - 7k²)/[√(k² + 1)³] = 0
k = 1/√7
S = 6*(1/√7)[√(1 - 3/7)]/(1 + 1/7) = (12/7)/(8/7)
= 3/2