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解:
(1)令x=0,y=0-1=-1,
∴點C的坐標(biāo)(0,-1);
(2)①平移后二次函數(shù)的解析式為y=-
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(x-2)
2+n,
由題意知:過C,A,B三點的圓的圓心一定在直線x=2上,點C為定點.
∴當(dāng)圓的半徑等于點C到直線x=2的距離時,圓的半徑最小,從而圓的面積最�。�
此時,圓的半徑為2,面積為4π.
設(shè)圓心為M,直線x=2與x軸交于點D,連接AM,則AM=2,
∵CM=2,OC=1,∴DM=1.
在Rt△AMD中,AD=
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=
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=
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,
∴點A的坐標(biāo)是(2-
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,0),代入拋物線得n=
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.
∴當(dāng)n=
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時,過C,A,B三點的圓的面積最小,最小面積為4π;
②如圖2,當(dāng)點P在直線y=
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x-1下方時,
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設(shè)直線y=
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x-1與x軸相交于點E,過點P作PN⊥EC于點N,PM∥y軸交EC于點M,則∠PMN=∠OCE,∠PNM=∠COE=90°,
∴△PMN∽△ECO,
∴
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,
令y=
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x-1=0.則x=
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,即OE=
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,CE=
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,
設(shè)點P的橫坐標(biāo)為m,則PM=MH+PH,
即PM=
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m-1+
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(m-2)
2-
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=
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(m
2-m-3),
∴PN=
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=
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(m
2-m-3),
根據(jù)題意,
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(m
2-m-3)=m,
解得m
1=3+2
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,m
2=3-2
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(不合題意,舍去),
即點P的坐標(biāo)是(3+2
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,-
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),
當(dāng)點P在直線y=
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x-1上方時,同理可得
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(m
2-m-3)=-m,
解得m
3=-
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-2(不合題意,舍去),m
4=
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-2,即點P的坐標(biāo)是(
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-2,2
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-5),
綜上,點P的坐標(biāo)是(3+2
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,-
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)或(
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-2,2
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-5).
分析:(1)由直線y=
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x-1與y軸交于點C,令x=0,求得y的值,即可求得點C的坐標(biāo);
(2)①首先設(shè)平移后二次函數(shù)的解析式為y=-
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(x-2)
2+n,由過C,A,B三點的圓的圓心一定在直線x=2上,點C為定點,即可得:當(dāng)圓的半徑等于點C到直線x=2的距離時,圓的半徑最小,從而圓的面積最小,則可求得n的值;
②分別從當(dāng)點P在直線AC下方時與當(dāng)點P在直線AC上方時去分析,借助于相似三角形的對應(yīng)邊成比例即可求得答案.
點評:此題考查了一次函數(shù)與坐標(biāo)軸交點的特點,二次函數(shù)的平移以及圓的性質(zhì)等知識.此題綜合性很強(qiáng),解題的關(guān)鍵是注意數(shù)形結(jié)合思想與分類討論思想的應(yīng)用.