两圆C1:x^2+y^2+4x-2y-5=0,C2:x^2+y^2-6x-4y-3=0的公共弦长为

两个方程相减
公共弦方程是10x+2y-2=0
y=-5x+1
代入C1
x^2+25x^2-10x+1+4x+10x-2-5=0
13x^2+2x-3=0
x1+x2=-2/13,x1*x2=-3/13
(x1-x2)^2=(x1+x2)^2-4x1*x2=160/169
(y-y2)^2=[(-5x1+1)-(-5x2+1)]^2=[-5(x1-x2)]^2=25*160/169
所以公共弦长=√[(x1-x2)^2+(y-y2)^2]
=4√260/13