C
分析:根據(jù)A點(diǎn)的坐標(biāo)為(
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,3)、AB=3BD,可以求得點(diǎn)D的坐標(biāo),從而得出反比例函數(shù)y=
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解析式,再根據(jù)A點(diǎn)坐標(biāo)得出AO直線解析式,進(jìn)而得出兩圖象的交點(diǎn)坐標(biāo),進(jìn)而得出AC的長(zhǎng)度,再利用直線與圓的位置關(guān)系得出答案.
解答:
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解:∵已知點(diǎn)A的坐標(biāo)為(
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,3),AB=3BD,
∴AB=3,BD=1,
∴D點(diǎn)的坐標(biāo)為(
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,1),
∴反比例函數(shù)y=
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解析式為:
y=
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,
∴AO直線解析式為:y=kx,
3=
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k,
∴k=
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,
∴y=
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x,
∴直線y=
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x與反比例函數(shù)y═
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x的交點(diǎn)坐標(biāo)為:
x=±1,
∴C點(diǎn)的橫坐標(biāo)為1,縱坐標(biāo)為
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,
過C點(diǎn)做CE垂直于OB于點(diǎn)E,
則CO=2,
∴AC=2
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-2,
∴CA的
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倍=
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(
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-1),
CE=
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,
∵
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(
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-1)-
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=
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-
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>0,
∴該圓與x軸的位置關(guān)系是相交.
故選:C.
點(diǎn)評(píng):此題主要考查了直線與圓的位置關(guān)系以及反比例函數(shù)的性質(zhì)以及直線與反比例函數(shù)交點(diǎn)坐標(biāo)的求法,綜合性較強(qiáng)得出AC的長(zhǎng)是解決問題的關(guān)鍵.