点（2a,a-1）在圆x平方+y平方-2y-4=0的内部,则a的取值范围是?A.-1

Question

点（2a,a-1）在圆x平方+y平方-2y-4=0的内部,则a的取值范围是?
A.-1

Answer 1

x² + y² - 2y - 4 = 0
圆心为(- D/2,- E/2) = (- 0/2,- (- 2)/2) = (0,1)
半径r = √[(D/2)² + (E/2)² - F] = √(0² + 1² - (- 4)) = √5
圆心与点(2a,a - 1)的距离是
d = √[(2a - 0)² + ((a - 1) - 1)²] = √(5a² - 4a + 4)
点(2a,a - 1)在圆内部,则d < r
√(5a² - 4a + 4) < √5
- 1/5 < a < 1
∴答案是D