Question

求圆C1；x²+y²-2x+2y-1=0与圆C2；x²+y²+2x-2y-3=0的公共弦长

求圆C1；x²+y²-2x+2y-1=0与圆C2；x²+y²+2x-2y-3=0的公共弦长

Answer 1

C1:x²+y²-2x+2y-1=0 (1)
C2:x²+y²+2x-2y-3=0 (2)
let C1, C2 intersect at P(x1,y1) ,Q(x2,y2)

(2)-(1)
4x-4y= 2
2x-2y =1 (3)
sub (3) into (1)
(1+2y)^2/4 + y^2 - (1+2y) +2y -1 =0
(1+2y)^2 + 4y^2 - 4(1+2y) +8y -4 =0
8y^2-4y-7 =0
y1+y2= 1/2
y1y2 = -7/8
(y1-y2)^2 = (y1+y2)^2 -4y1y2
= 1/4 + 7/2
=15/4
Similary
sub (3) into (1)
x^2 +(2x-1)^2/4 -2x + (2x-1) -1=0
4x^2 +(2x-1)^2 -8x + 4(2x-1) -4=0
8x^2-12x-7=0
x1+x2=3/2
x1x2=-7/8
(x1-x2)^2 = (x1+x2)^2 -4x1x2
=9/4 + 7/2
=23/4
公共弦长
=√[(x1-x2)^2+(y1-y2)^2]
=√(38/4)
=(1/2)√38

Answer 2

式子不会打，思路是：两圆方程相加和相减，得x y二元二次方程组，用消去法解得两个交点坐标，再用公式算两点距离