（1）已知椭圆C x^2/2+y^2=1 的右焦点为F .O为坐标原点 （1）求过点O,F并且与直线X=2相切的圆的方程

(1) c² = a² - b² = 2 - 1 = 1,c = 1
F(1,0)
显然圆心在x = (1 + 0)/2 = 1/2上,半径 = 2 - 1/2 = 3/2
圆心P(1/2,p):OP = 3/2 = √[(1/2 - 0)² + (p - 0)²]
p = ±√2
圆的方程:(x - 1/2)² + (y ±√2)² = 9/4
(2)显然B只能是上顶点或下顶点,设为上顶点(0,b),不妨设F(c,0)
BF斜率为-b/c,BC斜率为c/b
BC的方程:y = cx/b + b
y = 0,x = -b²/c
C(-b²/c,0)
CF的中垂线为x = (-b²/c + c)/2 = (c² - b²)/(2c)
FB的中点为N(c/2,b/2),中垂线为y - b/2 = (c/b)(x - c/2)
取x = (c² - b²)/(2c) ,y = 0
圆心M((c² - b²)/(2c),0)
MF = 半径2 = c - (c² - b²)/(2c)
c² + b² = a² = 4c (i)
离心率为c/a = 1/2 (ii)
由(i)(ii):c = 1,a = 2,b² = 3
椭圆的方程:x²/4 + y²/3 = 1