$P(1,1)$  is a fixed point on the circle $x^2+y^2-6x-4y+8=0$  and A, B are moving on the circumference of the same circle such that  $PA=PB=d(>0)$. The equation of the secant line AB, then $d$  is maximum is:

 When $d$ is maximum, A and B coincide to each other and coincide with $Q(5,3)$ the other end of the diameter through $P(1,1)$ . So secant $\overleftrightarrow{AB}$  coincides with tangent at $Q(5,3)$ .  
 Its equation is $5x+3y-3(x+5)-2(y+3)+8=0$ , i.e.$2x+y-13=0$,