Question

已知圆C:x∧2+y∧2+-2x-2y+1=0,直线L:y=kx,且与圆相交于P,Q两点,M(0,b),且MP⊥MQ

Answer 1

1、MP斜率为(1-y1)/(-x1),MQ斜率为(1-y2)/(-x2)

∵MP⊥MQ,∴(1-y1)/(-x1)*(1-y2)/(-x2)=-1 1-y1-y2+y1y2=-x1x2 式子1

将y=kx代入圆的方程 x²+k²x²-2x-2kx+1=0

得到x1x2=c/a=1/(1+k²),x1+x2=-b/a=(2+2k)/(1+k²)

因为y=kx 所以y1y2=kx1kx2=k²/(1+k²),y1+y2=(2k²+2k)/(1+k²)

代入式子1得 1-(2k²+2k)/(1+k²)+k²/(1+k²)=-1/(1+k²)

两边同乘以1+k²,解得 k=1

2、MP斜率为(b-y1)/(-x1),MQ斜率为(b-y2)/(-x2)

∵MP⊥MQ,∴(b-y1)/(-x1)*(b-y2)/(-x2)=-1 b²-b(y1+y2)+y1y2=-x1x2

代入得b²-b(2k²+2k)/(1+k²)+k²/(1+k²)=-1/(1+k²)

(b-1)²k²-2bk+b²+1=0