直线x+y-1=0被圆x²+y²-2x-2y-6=0所截得的弦长是?
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直线x+y-1=0被圆x²+y²-2x-2y-6=0所截得的弦长是?
直线x+y-1=0被圆x²+y²-2x-2y-6=0所截得的弦长是?
直线x+y-1=0被圆x²+y²-2x-2y-6=0所截得的弦长是?
(x-1)²+(y-1)²=9
圆心C(1,1),r=3
弦心距d=|1+1-1|/√(1²+1²)=√2/2
所以弦长=2√(r²-d²)=√34
因为x y-1=0,所以x=1-y
将x=1-y带入x^2 y^2-2x-2y-6=0
(1-y)^2 y^2-2(1-y)-2y-6=0
1-2y y^2 y^2-2 2y-2y-6=0
整理得2y^2-2y-7=0
解得:y1=(1 √15)/2,y2=(1-√15)
把y1、y2带入x=1-y得
x1=1-[(1 √15)/2]=(1-...
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因为x y-1=0,所以x=1-y
将x=1-y带入x^2 y^2-2x-2y-6=0
(1-y)^2 y^2-2(1-y)-2y-6=0
1-2y y^2 y^2-2 2y-2y-6=0
整理得2y^2-2y-7=0
解得:y1=(1 √15)/2,y2=(1-√15)
把y1、y2带入x=1-y得
x1=1-[(1 √15)/2]=(1-√15)/2
x2=1-[(1-√15)/2]= (1 √15)/2
所以直线与圆的两交点A[(1-√15)/2,(1 √15)/2]
B[(1 √15)/2,(1-√15)/2]
收起
用解方程,可两组(X1,Y1),(X2,Y2)的数值,再用两点之间的距离公式可计算弦长。