51问答网 > 求过直线2x+y+4=0和圆x^2+y^2+2x-4y+1=0的交点且面积最小的圆方程

求过直线2x+y+4=0和圆x^2+y^2+2x-4y+1=0的交点且面积最小的圆方程

过程谢谢
2024-12-26 03:31:25
推荐回答（2个）
回答1：

联立得：5x^2+26x+33=0
(x+3)(5x+11)=0
解得：x1=-3,x2=-11/5
y1=2, y2=2/5
两交点分别是A（-3，2）,B（-11/5，2/5）
所求面积最小的圆是以AB为直径的圆
圆心是AB中点（-13/5，6/5）
半径^2=1/4AB^2=1/4[(-3+11/5)^2+(2-2/5)^2]=4/5
面积最小的圆方程：(x+13/5)^2+(y-6/5)^2=4/5

回答2：

联立
2x+y+4=0
x^2+y^2+2x-4y+1=0
x1=-3，x2=-11/5
y1=2, y2=2/5
要过这两交点且面积最小，那么这两点的连线就是圆的直径
所以圆心：x0=（x1+x2）/2=-13/5,y0=(y1+y2)/2=6/5
半径：（1/2）*√(x1-x2)^2+(y1-y2)^2=4√5/5
圆方程：（x+13/5）^2+(y-6/5)^2=16/5

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