已知圆C:x2+y2-2x-2y+1=0,直线l:y=kx,且l与圆C相交于P、Q两点,点M(0,b),且MP⊥MQ.
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已知圆C:x2+y2-2x-2y+1=0,直线l:y=kx,且l与圆C相交于P、Q两点,点M(0,b),且MP⊥MQ.
(1)当b=1时,求k的值;
(2)求关于b和k的二元方程;
(3)求k的最小值.
(1)当b=1时,求k的值;
(2)求关于b和k的二元方程;
(3)求k的最小值.
(1)圆C:x2+y2-2x-2y+1=0可化为(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,
∵圆心C的坐标为(1,1),代入y=kx,
∴k=1;
(2)设P(x1,y1),Q(x2,y2),
由
x2+y2−2x−2y+1=0
y=kx,
得(1+k2)x2+2(k+1)x+1=0.
则x1+x2=−
2(k+1)
k2+1,x1x2=
1
1+k2.
∵MP⊥MQ,
∴kMP•kMQ=
y1−b
x1•
y2−b
x2=−1.
∵y1=kx1,y2=kx2,
∴-1=
y1−b
x1•
y2−b
x2=
(kx1−b)(kx2−b)
x1x2
=
k2x1x2−kb(x1+x2)+b2
x1x2
=k2−kb
x1+x2
x1x2+
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,
∵圆心C的坐标为(1,1),代入y=kx,
∴k=1;
(2)设P(x1,y1),Q(x2,y2),
由
x2+y2−2x−2y+1=0
y=kx,
得(1+k2)x2+2(k+1)x+1=0.
则x1+x2=−
2(k+1)
k2+1,x1x2=
1
1+k2.
∵MP⊥MQ,
∴kMP•kMQ=
y1−b
x1•
y2−b
x2=−1.
∵y1=kx1,y2=kx2,
∴-1=
y1−b
x1•
y2−b
x2=
(kx1−b)(kx2−b)
x1x2
=
k2x1x2−kb(x1+x2)+b2
x1x2
=k2−kb
x1+x2
x1x2+
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