(1)自点P作圆O的切线为y=kx-5k(x=5与圆O不相切)
圆O圆心(0,0),半径r=4
(-5k/(1+k^2))^2=16
k=正负4/3
所以y=4/3x-20/3或y=-4/3x+20/3与圆O相切
设切点为A
PA=根号(PO^2-OA^2)=根号(5^2-4^2)=3
(2)
设A(x1,y1),B(x2,y2),直线AB:y=kx-5k(x=5与圆O不相交)
y=kx-5k
x^2+y^2=16
=>(1+k^2)x^2-10k^2x+25k^2-16=0
x1+x2=10k^2/(1+k^2)
y1+y2=kx1-5k+kx2-5k
=k(x1+x2)-10k
=-10k/(1+k^2)
中点C(5k^2/(1+k^2),-5k/(1+k^2))