求以相交两圆C1:x^2+y^2+4x+1=0及C2:x^2+y^2+2x+2y+1=0的公共弦为直径的圆方程是?

题目

求以相交两圆C1:x^2+y^2+4x+1=0及C2:x^2+y^2+2x+2y+1=0的公共弦为直径的圆方程是?

答案

x^2+y^2+4x+1=0
x^2+y^2+2x+2y+1=0
2y-2x=0
y=x
2x^2+4x+1=0
2(x+1)^2=1
x1=-1+√2/2 y1=-1+√2/2
x2=-1-√2/2 y2=-1-√2/2
d^2=(x1-x2)^2+(y1-y2)^2=2*2=4
r^2=1
O'x=(x1+x2)/2=-1
O'y=(y1+y2)/2=-1
(x+1)^2+(y+1)^2=1