# Choose the correct statement about two

Question:

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

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

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

1. (1) circles have same centre

2. (2) circles have no meeting point

3. (3) circles have only one meeting point

4. (4) circles have two meeting points

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.