Question

已知定圆F1:x²+y²+10x+24=0,定圆F2:x²+y²-10x+9=0,动圆M与定圆F1,F2都外切

已知定圆F1:x²+y²+10x+24=0,定圆F2:x²+y²-10x+9=0,动圆M与定圆F1,F2都外切，求动圆圆心M的轨迹方程

Answer 1

F1:(x+5)^2+y^2=1 F2: (x-5)^2+y^2=4^2
F1圆心(-5,0), 半径1, F2圆心(5,0)半径4
设动圆圆心坐标(x,y)
√[(x+5)^2+y^2]-1=√[(x-5)^2+y^2]-4
√[(x+5)^2+y^2]=√[(x-5)^2+y^2]-3 两边平方

x^2+10x+25+y^2=x^2-10x+25+y^2+9-6√[(x-5)^2+y^2]
6√[(x-5)^2+y^2]=9-20x 两边平方

36x^2-360x+900+6y^2=81-360x+400x^2
364x^2-6y^2-819=0