直線mx-y+1-m=0即為y-1=m(x-1),過定點(1,1)；且m=(y-1)/(x-1)
又定點P(1,1)分弦AB為向量PB=2向量PA,則P,A,B三點必然共線
設(shè)P(X,Y)=P(1,1),A(X1,Y1),B(X2,Y2)
∵點A,B在圓C上,∴有X1²+(Y1-1)²=5,X2²+(Y2-1)²=5
由PB=2PA可得,PB²=4PA²
而PB²=(X2-X)²+(Y2-Y)²=(X2-1)²+(Y2-1)²=[X2²+(Y2-1)²]+1-2X2=6-2X2
PA²=(X1-X)²+(Y1-Y)²=(X1-1)²+(Y1-1)²=[X1²+(Y1-1)²]+1-2X1=6-2X1
∴有6-2X2=4(6-2X1),即3-X2=12-4X1,解得X2=4X1-9
由PB/PA=2可得,(X2-X)/(X-X1)=(X2-1)/(1-X1)=(4X1-9-1)/(1-X1)=2/1
解得X1=2,X2=-1
由于P,A,B共線,可得(Y2-Y)/(X2-X)=(Y-Y1)/(X-X1)=m
即(Y2-1)/(X2-1)=(1-Y1)/(1-X1)=m,代入X1,X2的值,解得Y2=3-2Y1
將X1,X2,Y1,Y2的值分別代入圓方程,得
X1²+(Y1-1)²=5,X2²+(Y2-1)²=5
兩式相減,得 X1²-X2²+[(Y1-1)²-(Y2-1)²]=0,即(X1-X2)(X1+X2)+(Y1+Y2-2)(Y1-Y2)=0
代入X1,X2,Y2的值,整理得 (Y1-1)²=1
∴Y1-1=±1,Y1=1±1,即Y1=0或2
∴當(dāng)Y1=0時,m=(1-Y1)/(1-X1)=(1-0)/(1-2)=-1,直線方程為：y=-x+2
當(dāng)Y1=2時,m=(1-Y1)/(1-X1)=(1-2)/(1-2)=1,直線方程為：y=x
∴直線l的方程為：y=-x+2或y=x
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