圆x^2+y^2+8x-4y=0与圆x^2+y^2=20关于直线y=2x+5对称,两圆交点为A,B求AB的长短问下最后一部为什么(1+4)(x1-x2)^2什么意思(把y=2x+5代入x^2+y^2=20得x^2+(2x+5)^2=20x^2+4x+1=0x1+x2=-4,x1x2=1(x1-x2)^2=(x1+x2)^2-4x1x2=16-4=12|AB

Question

圆x^2+y^2+8x-4y=0与圆x^2+y^2=20关于直线y=2x+5对称,两圆交点为A,B 求AB的长短
问下最后一部为什么(1+4)(x1-x2)^2什么意思
(把y=2x+5代入x^2+y^2=20得
x^2+(2x+5)^2=20
x^2+4x+1=0
x1+x2=-4,x1x2=1
(x1-x2)^2=(x1+x2)^2-4x1x2=16-4=12
|AB|^2=(1+4)(x1-x2)^2=5*12=60)

Answer 1

|AB|^2=（y1-y2)^2+(x1-x2)^2=(2x1+5-2x2-5)^2+(x1-x2)^2={2(x1-x2)}^2+(x1-x2)^2=(4+1)(x1-x2)^2=60