已知圆c经过坐标原点,且与直线x-y+2=0相切,切点为A(2.4)1:求圆c的方程;2:若斜率为-1的直线L与圆c相交与两个不同的点M.N,求AM*AN的取值范围(AM.AN上有箭头).

Question

已知圆c经过坐标原点,且与直线x-y+2=0相切,切点为A(2.4)
1:求圆c的方程;
2:若斜率为-1的直线L与圆c相交与两个不同的点M.N,求AM*AN的取值范围(AM.AN上有箭头).

Answer 1

设圆C的方程为
(x-a)^2 + (y-b)^2 = r^2,
则由圆C过原点,有
a^2 + b^2 = r^2,
(x-a)^2 + (y-b)^2 = a^2 + b^2,
x^2 -2ax + y^2 -2by = 0.
A(2,4)在圆上,有
0 = 2^2 - 2a*2 + 4^2 - 2b*4 = 4 - 4a + 16 - 8b = 4[5-a-2b].
5=a+2b,a=5-2b.
x^2 -2(5-2b)x +y^2-2by=0.
直线x-y+2=0与圆C相切于切点A(2.4),有
0 = 2x -2(5-2b) + 2y*y' - 2b = 2[x+y*y'-5+b].
y'(2)=1.y(2)=4.
0 = 2[2 + 4*1 -5 + b]=2(1+b),
b=-1,
a=5-2b=5+2=7,
r^2=a^2+b^2=7^2 + 1=50.
圆C的方程为,
(x-7)^2 + (y+1)^2 = 50.
设直线L的方程为
y=t-x.t为常数.
50 = (x-7)^2 + (t-x+1)^2 = (x-1)^2 -12(x-1) + 36 + t^2 -2t(x-1) + (x-1)^2 = 2(x-1)^2 - 2(x-1)[6+t] + t^2 + 36,
方程0 = (x-1)^2 - (x-1)(6+t) + t^2/2 - 7有2个不同的实根.
0 t^2 - 12t - 64 = (t-16)(t+4),
-4