Question

一圆经过点P(2,-1)和直线x-y-1=0相切,且圆心在直线2x+y=0上,求该圆的方程

Answer 1

解：可设圆心C(t,-2t).由题设得√[(t-2)²+(2t-1)²]=|3t-1|/√2=R.解得t=1,或t=9.当t=1时，圆心(1,-2),半径R=√2.圆的方程为(x-1)²+(y+2)²=2.当t=9时，圆心(9,-18),半径R=13√2.圆的方程为(x-9)²+(y+18)²=338.

Answer 2

圆心在直线2x+y=0
圆心 = (m , -2m)
圆心和直线x-y-1=0 的距离
=|m+2m-1|/√2
=|3m-1|√2
=r
圆的方程
(x-m)^2+(y+2m)^2 =(3m-1)^2/2
过点P(2,-1)
(2-m)^2+(-1+2m)^2 =(3m-1)^2/2
5m^2 -8m +5 =(3m-1)^2/2
10m^2-16m+10 =9m^2- 6m+1
m^2-10m+9=0
(m-1)(m-9) =0
m=1 or 9

圆的方程
(x-1)^2+(y+2)^2 =2 or
(x-9)^2+(y+18)^2 =338

Answer 3

设圆方程为（x-a)^2+(y-b)^2=r^2,则依题意得2a+b=0,(2-a)^2+(-1-b)^2=r^2,1/2(a-b-1)^2=r^2,联立三个式子解得a=1,b=-2,r^2=2或a=9,b=-18,r^2=338.带入所设圆方程即可