Question

Choose the correct statement about two circles whose equations are given below:
$x^2+y^2-10x-10y+41=0$
$x^2+y^2-22x-10y+137=0$

• A. circles have no meeting point
• B. circles have two meeting point
• C. circles have only one meeting point
• D. circles have same centre

Answer 1

The correct option is C circles have only one meeting point

Let $S_1:x^2+y^2-10x-10y+41=0$
$\Rightarrow (x-5{)}^2+(y-5{)}^2=9$
Centre $\left(C_1\right)=(5,5)$
Radius $r_1=3$

$S_2:x^2+y^2-22x-10y+137=0$
$\Rightarrow (x-11{)}^2+(y-5{)}^2=9$
Centre $\left(C_2\right)=(11,5)$
Radius $r_2=3$

Distance $\left(C_1{C}_2\right)=\sqrt{(5-11{)}^2+(5-5{)}^2}$
Distance $\left(C_1{C}_2\right)=6$
$\because r_1+r_2=3+3=6$
$\therefore$ Circles touch externally.
Hence, circle have only one meeting point.