Question

求圆心在直线X-Y-4=0上,并且经过X^2+y^2-4x-3=0与x^2+y^2-4y-3=0的点的圆的方程

Answer 1

求圆心在直线X-Y-4=0上,并且经过两园X²+y²-4x-3=0与x²+y²-4y-3=0的交点的圆的方程

解：设过两园交点的园的方程为
X²+y²-4x-3+λ(x²+y²-4y-3)=(1+λ)x²+(1+λ)y²-4x-4λy-3(1+λ)
=(1+λ)[x²-4/(1+λ)]+(1+λ)[y²-4λ/(1+λ)]-3(1+λ)
=(1+λ)[(x-2/(1+λ)]²-4/(1+λ)+(1+λ)[y-2λ/(1+λ)]²-4λ²/(1+λ)-3(1+λ)
=(1+λ)[(x-2/(1+λ)]²+(1+λ)[y-2λ/(1+λ)]²-4/(1+λ)-4λ²/(1+λ)-3(1+λ)
=(1+λ){[(x-2/(1+λ)]²+[y-2λ/(1+λ)]²-[(4λ²+4)/(1+λ)²+3]}=0
即得[(x-2/(1+λ)]²+[y-2λ/(1+λ)]²-[(4λ²+4)/(1+λ)²+3]=0.............(1)
园心横坐标为2/(1+λ)，纵坐标为2λ/(1+λ)；园心在直线x-y-4=0上，因此有等式：
2/(1+λ)-2λ/(1+λ)-4=0；去分母得 2-2λ-4(1+λ)=-6λ-2=0，故λ=-1/3.
代入(1)式即得园的方程为：
(x-3)²+(y+1)²-13=0