题目内容
已知点
是直线
上一动点,
是圆C:
的两条切线,A、B是切点,若四边形
的最小面积是2,则
的值为?
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试题分析:利用切线的性质,建立四边形PACB的面积与切线长PA的关系式,根据四边形PACB面积的最小值可以得到PA的最小值,再利用PA与CP之间的关系可以得到CP的最小值,而CP的最小值即圆心C到直线的距离,从而可以建立关于k的方程求得k的值.
C:
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如图,∵
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∵
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又∵
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即点C到直线的距离为
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∴
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解得:
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∴
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