设动点P在直线l:y=x+4上运动,过点P作圆O:x^2+y^2=1的两条切线PA,PB,其中A,B为切点,

这题目是考奥林匹克高手的,只给5分,太少了呀.
P(a,4+a)
A(x,y)
圆O:x^2+y^2=1.(1)
O(0,0),r=1
PA^2=PB^2=PO^2-r^2=a^2+(4+a)^2-1
圆P:(x-a)^2+([y-(4+a)]^2=PA^2=PB^2=a^2+(4+a)^2-1.(2)
(1)-(2):
直线AB:ax+(4+a)y=1.(3)
k(AB)=-a/(4+a)
M(x,y)
k(OP)=(4+a)/a=y/x
a=4x/(y-x),代入(3)
[4x/(y-x)]*x+[4+4x/(y-x)]y=1
4x^2+4y^2+x-y=0
(x+0.125)^2+(y-0.125)^2=1/32