Question

已知⊙O1：x²+y²+2x-6y+1=0与⊙O2：x²+y²-4x+2y-11=0，

求两圆的公共弦所在的直线方程及公共弦长。

Answer 1

x²+y²+2x-6y+1=0 (1）
x²+y²-4x+2y-11=0 （2）

(1)-(2)
6x-8y+12 =0
x = (4y-6)/3 (3)
Sub (3) into (1)
[(4y-6)/3]^2+ y^2 + 2[(4y-6)/3] - 6y + 1 =0
16y^2-48y+36 + 9y^2 + 24y-36 - 54y + 9 =0
25y^2-78y+9 =0
(y-3)(25y-3) =0
y = 3 or 3/25
when y = 3 , x=2
when y = 3/25, x = (4(3/25)-6)/3 = -46/25

直线方程
(y-3)/(x-2) = (3/25 - 3)/ (-46/25 - 2)
= 72/96
= 3/4
4(y-3) = 3(x-2)
3x-4y+6 = 0
公共弦长
=√ {(-46/25 - 2)^2 + (3/25 - 3)^2}
=120/25
= 24/5