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 the same center.

Answer 1

The correct option is C

Circles have only one meeting point.

Explanation for the correct option:

Step$1$. To find the center and radius of the first circle:

The given circle $x^2+y^2–10x–10y+41=0$

$$\begin{gathered} \Rightarrow x^2-10x+25+y^2-10y+25=9 \\ \Rightarrow {\left(x-5\right)}^2+{\left(y-5\right)}^2=3^2 \end{gathered}$$

Thus center $C_1(5,5)$and radius $r_1=3$

Step$2$. To find the center and radius of the second circle:

The given circle $x^2+y^2–22x–10y+137=0$

$$\begin{gathered} \Rightarrow x^2-22x+121+y^2-10y+25=9 \\ \Rightarrow {\left(x-11\right)}^2+{\left(y-5\right)}^2=3^2 \end{gathered}$$

Thus center $C_2(11,5)$and radius $r_2=3$

Step$3$. To find the distance between centers of circles:

$C_1{C}_2=\sqrt{{\left(11-5\right)}^2+{\left(5-5\right)}^2}=6$

and $r_1+r_2=3+3=6$

Since $C_1{C}_2=r_1+r_2$

So, the circles touch externally.

Thus the circle has only one meeting point.

Hence the correct option is C.