Question

如图,圆x²+y²=8内有一点P（-1,2）,AB为过点P且倾斜角为α的弦.

Answer 1

1）α=135°时,直线AB的斜率为k=tan135°=-1

又AB过点P(-1,2),∴直线AB方程为y=-1(x+1)+2=-x+1

将直线代入圆方程得 x^2+(-x+1)^2=8

整理得 2x^2-2x-7=0

∴x1+x2=1,x1x2=-7/2; y1+y2=-(x1+x2)+2=1,

y1y2=(-x1+1)(-x2+1)=-x1x2-(x1+x2)+1=7/2-1+1=7/2

∴AB=√[(x1-x2)^2+(y1-y2)^2]

=√[(x1+x2)^2-4x1x2+(y1+y2)^2-4y1y2]

=√[1^2+4*7/2+1^2-4*7/2]

=√2

2）当弦AB被点P平分时,必有AB⊥OP

而k(OP)=2/(-1)=-2,∴k(AB)=-1/k(OP)=1/2

又直线AB过点P(-1,2)

∴直线AB方程为 y=1/2*(x+1)+2=(x+5)/2