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解:(1)連MA,MB,如圖:
∵MA=MB OM⊥AB∠AMB=120°,
∴∠BMO=
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∠AMB=60°,
∴∠OBM=30°,
∴OM=
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MB=1,
∴M(0,1);
(2)∵OC=MC-MO=1 OB=
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=
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,
∴C(0,-1)B(
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,O),
∵經(jīng)過A,B,C三點的拋物線關(guān)于y軸對稱,
∴設(shè)經(jīng)過A,B,C三點的拋物線的解析式為y=ax
2+c,
把C(0,-1)和(
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,0)分別代入上式,
得:a=
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,c=-1,
∴y=
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x
2-1;
(3)連接AM并延長交圓于點P,連接PB,
∵90°的圓周角對的弦是直徑,
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∴∠P≠90°,
∴∠B=90°或∠A=90°,
當∠B=90°時,AP是直徑,
∵弦AB所對的圓心角為120度,
∴∠P=60°,
∴∠A=30°,
∵圓的半徑為2cm,
∴AP=4,
∴BP=2,
∴點P的坐標為(
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,2),
同理可得:當∠A=90°時,點P的坐標為(-
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,2).
∴點P的坐標為(
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,2),(-
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,2).
分析:(1)連接MA,MB,根據(jù)等腰三角形的性質(zhì)可知∠AMO=
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AMB=60°,由直角三角形的性質(zhì)可求出M點的坐標.
(2)根據(jù)△AOM與△BOM是直角三角形,∠AMO=∠BMO=60°,可求出A、B兩點的坐標,因為A、B兩點關(guān)于y軸對稱,故此拋物線關(guān)于y軸對稱,根據(jù)此特點可設(shè)出拋物線的解析式,把A、B兩點的坐標代入即可求出未知數(shù)的值,從而求出其解析式.
(3)設(shè)P(m,n),根據(jù)P在圓上列出方程及PA
2+PB
2=AB
2即可求解.
點評:本題考查的是圓的性質(zhì)及二次函數(shù)圖象上點的坐標特點,比較復(fù)雜,但難度適中,注意細心運算.