△ABC中,∠C=90°,AC=3,CB=6,若以C為圓心,以r為半徑作圓,那么:
(1)當(dāng)直線AB與⊙C相離時(shí),r的取值范圍是 ;
(2)當(dāng)直線AB與⊙C相切時(shí),r的取值范圍是 ;
(3)當(dāng)直線AB與⊙C相交時(shí),r的取值范圍是 .
【答案】
分析:根據(jù)勾股定理,求出AB,再根據(jù)面積求得點(diǎn)C到AB的距離d,當(dāng)r<d時(shí),直線AB與⊙C相離;當(dāng)r=d時(shí),直線AB與⊙C相切;當(dāng)r>d時(shí),直線AB與⊙C相交.
解答:解:∵∠C=90°,AC=3,CB=6,∴AB=3
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,
∴3×6=3
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d,∴d=
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,
∴(1)當(dāng)直線AB與⊙C相離時(shí),r的取值范圍是 r<
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;
(2)當(dāng)直線AB與⊙C相切時(shí),r的取值范圍是 r=
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;
(3)當(dāng)直線AB與⊙C相交時(shí),r的取值范圍是 r>
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.
故答案為r<
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;r=
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;r>
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.
點(diǎn)評:本題考查了直線和圓的位置關(guān)系、三角形的面積和勾股定理,
點(diǎn)到的距離d,當(dāng)r<d時(shí),直線AB與⊙C相離;當(dāng)r=d時(shí),直線AB與⊙C相切;當(dāng)r>d時(shí),直線AB與⊙C相交.