求通过直线2x-y+3=0与圆x2+y2+2x-4y+1=0的交点,且面积为最小的圆的方程.

该圆以交点距离为直径,两个交点中点为圆心.
将直线方程代入圆方程,分别消去x或者y.
x^2+(2x+3)^2+2x-4(2x+3)+1=0
=>5x^2+6x-2=0
=>x1+x2=-3/5
x1x2=-2/5
|x2-x1=根号[(x1+x2)^2-4x1x2]=7/5
(y-3)^2/4+y^2+(y-3)-4y+1=0
=>5y^2/4-9y/2+1/4=0
=>5y^2-18y+1=0
=>y1+y2=9/5
y1y2=1/5
|y1-y2|=（根号61）/5
直径=根号[(x2-x1)^2+(y2-y1)^2]=（根号110）/5
圆心((x1+x2)/2,(y1+y2)/2)=(-3/10,9/10)
(x+3/10)^2+(y-9/10)^2=110/25=22/5