Question

求圆心在直线2x-y-3=0上,过(3,-2)(5,2)圆的方程 若y=2/x+m有公共点,求m取值范围

Answer 1

2x-y-3=0

y=2x-3

C(a,2a-3)

r^2=(a-3)^2+(2a-3+2)^2=(a-5)^2+(2a-3-2)^2

a=2

C(2,1),r^2=10

C:(x-2)^2+(y-1)^2=10

y=(x/2)+m

x-2y+2m=0

d^2=|2-2*1+m|^2/5≤r^2

-5√2≤m≤5√2