已知a,b∈R,⊙C1:x2+y2-4x+2y-a2+5=0与⊙C2:x2+y2-(2b-10)x-2by+2b2-10b+16=0交于不同两点A(x1,y

2025-05-09 06:30:25
推荐回答(1个)
回答1:

由题意,将A(x1,y1),B(x2,y2)分别代入圆的方程,两方程相减可得:
-4x1+2y1+4x2-2y2=0,-(2b-10)x1-2by1+(2b-10)x2+2by2=0
∴y1-y2=2(x1-x2)①,2b(y1-y2)=-(2b-10)(x1-x2)②
①代入②可得4b(x1-x2)=-(2b-10)(x1-x2
∵x1≠x2,∴4b=-2b+10
∴b=

5
3

故答案为:
5
3