Question

数学题，急求答案！！！！！！

数学题，急求答案！！！！！！园c1:(x-1)²＋y²＝4与园c2:(x+1)²＋(y-3)²＝9相交弦所在直线l，则l被园o:x²+y²＝4截得的弦长为多少？

Answer 1

相交弦的方程为（x +1）^2+(y -3)^2-(x -1)^2-y ^2=9-4,
4x -6y +4=0, 或2x -3y +2=0,
与圆x^2+y^2=4联立方程求得交点(x1,y1),(x2,y2),y=2(x+1)/3,
x^2+4(x+1)^2/9=4, 13x^2+8x-32=0,
(x1-x2)^2=(x1+x2)^2-4x1x2=64/169+128/13=64×27/169,
(y1-y2)^2={2(x1+1)/3-2(x2+1)/3}^2=4(x1-x2)^2/9=64×4×3/169,
弦长=√{(x1-x2)^2+(y1-y2)^2}=8√39/13