Question

已知圆c的圆心在直线x-y-4=0上,并且经过两圆c1:x^2+y^2-4x-3=0

Answer 1

C1:x^2+y^2-4x-3=0

C2:x^2+y^2-4y-3=0

两式相减

得交点弦:x=y

x=y代入x^2+y^2-4x-3=0

解得x=(2±√10)/2

则y=x=(2±√10)/2

交点弦中点坐标(1,1)

交点弦中垂线过圆心C

中垂线:y-1=-1(x-1) x+y-2=0

与x-y-4=0交点C(3,-1)

半径√([(2+√10)/2-3]^2+[(2+√10)/2+1]^2)=√13

C:(x-3)^2+(y+1)^2=13

参考图:http://hi.baidu.com/%EE%A8%F4%D4/album/item/cc75f035cffbaa97d0a2d3a8.html