以點(diǎn)P(3,0)為端點(diǎn),與圓x2+y2=1相切的切線段的長(zhǎng)為________.
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分析:根據(jù)題意畫出圖形,得到線段PQ為所求的切線段長(zhǎng),由切線的性質(zhì),圓的切線垂直于過切點(diǎn)的直徑,得到三角形OPQ為直角三角形,根據(jù)P的坐標(biāo)和圓的半徑分別求出|OP|和|OQ|,利用勾股定理即可求出|PQ|的長(zhǎng),即為所求的切線段長(zhǎng).
解答:
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解:根據(jù)題意畫出圖形,如圖所示:
過點(diǎn)P作圓O的切線PQ,切點(diǎn)為點(diǎn)Q,連接OQ,
∴PQ⊥OQ,由圓的方程得到:圓心O坐標(biāo)為(0,0),半徑OQ=1,
在直角三角形OPQ中,|OQ|=1,|OP|=3,
根據(jù)勾股定理得:|PQ|=
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=2
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,
則以點(diǎn)P為端點(diǎn),與圓相切的切線段的長(zhǎng)為2
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.
故答案為:2
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點(diǎn)評(píng):此題要求學(xué)生掌握直線與圓相切時(shí)滿足的性質(zhì),考查了數(shù)形結(jié)合的數(shù)學(xué)思想.學(xué)生往往借助圖形來解答此類題,直觀形象,有利于更好的解題.