Choose the correct statement about two circles whose equations are given below :

$x^{2}+y^{2}-10 x-10 y+41=0$

$\mathrm{A}(5,5), \mathrm{R}_{1}=3$

$x^{2}+y^{2}-22 x-10 y+137=0$

$\mathrm{B}(11,5), \mathrm{R}_{2}=3$

$\mathrm{AB}=6=\mathrm{R}_{1}+\mathrm{R}_{2}$

Touch each other externally

$\Rightarrow$ circles have only one meeting point.