Question:
Choose the correct statement about two circles whose equations are given below:
$x^{2}+y^{2}-10 x-10 y+41=0$
$x^{2}+y^{2}-22 x-10 y+137=0$
Correct Option: , 3
Solution:
$x^{2}+y^{2}-10 x-10 y+41=0$
$\mathrm{~A}(5,5), \mathrm{R}_{1}=3$
$x^{2}+y^{2}-22 x-10 y+137=0$
$\mathrm{B}(11,5), \mathrm{R}_{2}=3$
$\mathrm{AB}=6=\mathrm{R}_{1}+\mathrm{R}_{2}$
Touch each other externally
$\Rightarrow$ circles have only one meeting point.