与方程(1+λ)x²+(1+λ)y²+6x-8y-25(λ+3)=0所表示的所有圆都相切的直线方程
(1+λ)[x²+6x/(1+λ)]+(1+λ)[y²-8/(1+λ)]-25(λ+3)=0
(1+λ)[x+3/(1+λ)]²+(1+λ)[y-4/(1+λ)]²-25/(1+λ)-25(λ+3)=0
[x+3/(1+λ)]²+[y-4/(1+λ)]²-25/(1+λ)²-25(λ+3)/(1+λ)=0
[x+3/(1+λ)]²+[y-4/(1+λ)]²=[5(λ+2)/(1+λ)]²
圆心(-3/(1+λ),4/(1+λ)),半径R=5(λ+2)/(1+λ)
由于[4/(1+λ)/[-3/(1+λ)]=-4/3=常量,∴所有圆的圆心都在过原点的直线y=-(4/3)x上.
圆心到原点的距离d=5/(1+λ),R-d=5(λ+2)/(1+λ)-5/(1+λ)=(5λ+5)/(1+λ)=5=常量,
故所有圆都内切于点(3,-4),∴与所有圆都相切的公切线方程为：y=(3/4)(x-3)-4=(3/4)x-25/4