Question

高一数学圆的题

已知圆C经过点M（1，3），且圆心在直线y=x+1上，（1）若圆C与直线L:x-2y-3=0相切，求圆C的方程（2）若原点O始终在圆C内，求圆C的面积的取值范围。

Answer 1

设圆心坐标为C(a, b), b = a + 1, C(a, a +1)
圆C经过点M（1，3）: (1 -a)^2 + (3 - a -1)^2 = r^2
2a^2 -6a +5 = r^2 (1)
1. 圆C与直线L:x-2y-3=0相切, C与直线L距离(d)等于半径.
d^2 = |a - 2(a+1) -3|^2/(1 + 2^2) = (a+5)^2/5
(a+5)^2/5 = 2a^2 -6a +5
a(9a-40) = 0
a = 0, a = 40/9
A: a = 0:
r^2 = 5
圆心C(0, 1)
圆C的方程: x^2 + (y-1)^2 = 5
B: a = 40/9
r^2 = 1445/81
圆心C(40/9, 49/9)
圆C的方程: (x-40/9)^2 + (y-49/9)^2 = 1445/81

2. 若原点O始终在圆C内, OC < CM, OC^2 < CM^2
OC^2 = (a - 0)^2 + (a+1 -0)^2 = a^2 + (a+1)^2
CM^2 = (a - 1)^2 + (a + 1 -3)^2 = (a-1)^2 + (a-2)^2
a^2 + (a+1)^2 < (a-1)^2 + (a-2)^2
8a < 4
a < 1/2
r^2 = 2a^2 -6a +5 = 2(a - 3/2) +1/2
r^2是以(3/2. 1/2)为顶点，开口向上的抛物线. a< 1/2时, 抛物线为左半部分.
圆C的面积S: πr^2 > π(1/2)^2 = π/4
π/4 < S < +∞