已知圆C:x2+y2-2x-2y+1=0,直线l:y=kx,且l与C相交于P、Q两点,点M(0,b),且MP⊥MQ.(Ⅰ)当b=1时

2025-05-08 20:30:00
推荐回答(1个)
回答1:

(Ⅰ)圆C:(x-1)2+(y-1)2=1,当b=1时,点M(0,b)在圆C上,
当且仅当直线l经过圆心C时,满足MP⊥MQ.…(2分)
∵圆心C的坐标为(1,1),∴k=1.…(4分)
(Ⅱ)由

y=kx
x2+2?2x?2y+1=0
,消去y得:(1+k2)x2-2(1+k)x+1=0.①
设P(x1,y1),Q(x2,y2),
x1+x2
2(1+k)
1+k2
x1x2
1
1+k2
.…(6分)
∵MP⊥MQ,∴
MP
?
MQ
=0

∴(x1,y1-b)?(x2,y2-b)=0,即 x1x2+(y1-b)(y2-b)=0.
∵y1=kx1,y2=kx2
∴(kx1-b)(kx2-b)+x1x2=0,即(1+k2)x1x2?kb(x1+x2)+b2=0.…(8分)
(1+k2)?
1
1+k2
?kb?
2(1+k)
1+k2
+b2=0
,即
2k(1+k)
1+k2
b2+1
b
=b+
1
b

f(b)=b+