已知矩形ABCD的对角线交于点P(2,0),边AB所在直线的方程为x-3y-6=0,点(-1,1)在边AD所在的直线上,
(1)求矩形ABCD的外接圆的方程;
(2)已知直线l:(1-2k)x+(1+k)y-5+4k=0(k∈R),求证:直线l与矩形ABCD的外接圆恒相交,并求出相交的弦长最短时的直线l的方程.
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