过动点P(x,y)引圆(x-1)^2+(y-2)^2=1的切线,其切线长等于点P到原点O的距离

过动点P(x,y)引圆(x-1)^2+(y-2)^2=1的切线,其切线长等于点P到原点O的距离
(1)求点P的坐标值（2）求切线长最短时,P点的坐标.
(1)设P点坐标为P(m,n),切点为T,圆(x-1)^2+(y-2)^2=1的圆心为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
因为|PO|=|PT|,所以|PO|^2=|PT|^2,即m^2+n^2=m^2+n^2-2m-4n+4,化简得m+2n-2=0,
即动点P在直线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
当n=4/5时,|PT|^2取得最小值4/5,此时切线PT长最短,|PT|min=2√5/5,
此时m=2-2n=2/5,
即切线长最短时,P点的坐标为(2/5,4/5).