Let equation of circle touches the positive coordinate axes having radius $r$ is
$(x-r)^2+(y-r)^2=r^2$
$\because$ The circle (i) touches the another given circle $x^2+y^2-12 x-10 y+52=0$ externally then
$c_1 c_2=r_1+r_2$
Where $c_1(r, r), c_2(6,5), r_1=r$ and $r_2=3$
$\therefore \quad \sqrt{(r-6)^2+(r-5)^2}=r+3$
$\Rightarrow \quad 2 r^2-22 r+61=r^2+6 r+9$
$\Rightarrow \quad r^2-28 r+52=0 \Rightarrow(r-26)(r-2)=0$
$\Rightarrow \quad r=2$ or 26
$\therefore$ There are two possible circles having centres $(2,2)$ and $(26,26)$ respectively.
So, required distance between centres
$\sqrt{(6-2)^2+(5-2)^2} \text { or } \sqrt{(26-6)^2+(26-5)^2}$
$=5$ or 29.