题目内容
已知点P(x,y)是直线kx+y+4=0(k>0)上一动点,PA,PB是圆C:x2+y2-2y=0的两条切线,A,B为切点,若四边形PACB的最小面积是2,则k的值为( ).
A.4 | B.3 | C.2 | D.![]() |
C
圆C的方程可化为x2+(y-1)2=1,因为四边形PACB的最小面积是2,且此时切线长为2,故圆心(0,1)到直线kx+y+4=0的距离为
,即
=
,解得k=±2,又k>0,所以k=2.
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