若P(2,-1)为圆(x-1)2+y2=25的弦AB的中点，则直线AB的方程是A.X-Y-3=0 B.2X+Y-3=0 C.X+Y-1=0 D.2X-Y-5=0
热心网友
设A(x1,y1),B(x2,y2)所以(x1-1)^2+y1^2=25,(x2-1)^2+y2^2=25相减得:(x1+x2-2)(x1-x2)+(y1+y2)(y1-y2)=0又因为AB中点为P(2,-1),所以(x1+x2)/2=2,(y1+y2)/2=-1所以x1+x2=4,y1+y2=-2,代入上式得:2(x1-x2)-(y1-y2)=0所以(y1-y2)/(x1-x2)=2,即AB斜率k=2所以直线AB方程为y+1=2(x-2),即2x-y-5=0