Let the circumcenter of a triangle vertices  $A(a,3),B(b,5)$  and  $C(a,b),ab>0$  be  $P(1,1)$ . If the line AP intersects the line BC at the point  $Q(k_1,k_2)$ , then  $k_1+k_2$  is equal to 

P is circumcenter.
 $$\begin{gathered} \therefore A{P}^2=B{P}^2=C{P}^2 \\ \Rightarrow {(a-1)}^2+{(3-1)}^2={(b-1)}^2+{(5-1)}^2={(a-1)}^2+{(b-1)}^2 \\ \Rightarrow {(a-1)}^2+4={(b-1)}^2+16={(a-1)}^2+{(b-1)}^2 \end{gathered}$$
$\Rightarrow {(b-1)}^2=4$   and  $\Rightarrow {(a-1)}^2=16$
If b=3 and a=5 (as ab>0), then 
$A\equiv (5,3),B\equiv (3,5)$   and  $C\equiv (5,3)$
This is not possible.
So, b=-1 and a=-3 (as ab>0)
 $A\equiv (-3,3),B\equiv (-1,5)$  and  $C\equiv (-3,-1)$
So, equation of BC is  $3x-y+8=0$ ………………….. (1)
Also, equation of AP is  $x+2y-3=0$ ………………. (2)
Solving (1) and (2) we get  $(k_1,k_2)\equiv (-\frac{13}{7},\frac{17}{7})$
$k_1+k_2=\frac{-13+17}{7}=\frac{4}{7}$ .