過動(dòng)點(diǎn)P(x,y)引圓(x-1)^2+(y-2)^2=1的切線,其切線長等于點(diǎn)P到原點(diǎn)O的距離
(1)求點(diǎn)P的坐標(biāo)值（2）求切線長最短時(shí),P點(diǎn)的坐標(biāo).
(1)設(shè)P點(diǎn)坐標(biāo)為P(m,n),切點(diǎn)為T,圓(x-1)^2+(y-2)^2=1的圓心為Q,
|PO|^2=m^2+n^2,|PT|^2=|PQ|^2-|QT|^2=(m-1)^2+(n-2)^2-1=m^2+n^2-2m-4n+4
因?yàn)閨PO|=|PT|,所以|PO|^2=|PT|^2,即m^2+n^2=m^2+n^2-2m-4n+4,化簡(jiǎn)得m+2n-2=0,
即動(dòng)點(diǎn)P在直線x+2y-2=0上.
（2）由（1）得,
|PT|^2=m^2+n^2-2m-4n+4
=(2-2n)^2+n^2-2(2-2n)-4n+4
=5n^2-8n+4
當(dāng)n=4/5時(shí),|PT|^2取得最小值4/5,此時(shí)切線PT長最短,|PT|min=2√5/5,
此時(shí)m=2-2n=2/5,
即切線長最短時(shí),P點(diǎn)的坐標(biāo)為(2/5,4/5).