题目

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.

提问时间：2020-12-23

答案

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 < +∞