解析几何,圆和直线已知⊙C:x²+(y-1)²=5,直线l:mx-y+1-m=0

解析几何,圆和直线已知⊙C:x²+(y-1)²=5,直线l:mx-y+1-m=0

直线mx-y+1-m=0即为y-1=m(x-1),过定点(1,1)；且m=(y-1)/(x-1)又定点P(1,1)分弦AB为向量PB=2向量PA,则P,A,B三点必然共线设P(X,Y)=P(1,1),A(X1,Y1),B(X2,Y2)∵点A,B在圆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共线,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将X1,X2,Y1,Y2的值分别代入圆方程,得X1²+(Y1-1)²=5,X2²+(Y2-1)²=5两式相减,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∴当Y1=0时,m=(1-Y1)/(1-X1)=(1-0)/(1-2)=-1,直线方程为：y=-x+2当Y1=2时,m=(1-Y1)/(1-X1)=(1-2)/(1-2)=1,直线方程为：y=x∴直线l的方程为：y=-x+2或y=x希望对你有帮助

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