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 points
C. circles have only one meeting point
D. circles have the same centre

Answer 1

Hint: In this question, we need to figure out how many meeting points two circles have. We must first find the radius and centre of each equation of the circle. If the distance between their centres is equal to the total of their radii then these two circles will be touching externally and have a single meeting point.

Complete step-by-step solution:
Consider \[{x^2} + {y^2} - 10x - 10y + 41 = 0\]
\[
  {x^2} + {y^2} - 10x - 10y + 41 = 0 \\
   \Rightarrow {\left( {x - 5} \right)^2} + {\left( {y - 5} \right)^2} = 9 \\
   \Rightarrow {\left( {x - 5} \right)^2} + {\left( {y - 5} \right)^2} = {\left( 3 \right)^2}
 \]
The standard equation of a circle having centre \[\left( {h,k} \right)\] and radius \[r\] is \[{\left( {x - h} \right)^2} + {\left( {x - k} \right)^2} = {r^2}\]
By comparing, we get
Thus, the centre of a circle is \[{C_1}\left( {h,k} \right) = {C_1}\left( {5,5} \right)\]
Also, the radius of a circle is \[{r_1} = 3\]
Now, consider \[{x^2} + {y^2} - 22x - 10y + 137 = 0\]
\[
  {x^2} + {y^2} - 22x - 10y + 137 = 0 \\
   \Rightarrow {\left( {x - 11} \right)^2} + {\left( {y - 5} \right)^2} = 9 \\
   \Rightarrow {\left( {x - 11} \right)^2} + {\left( {y - 5} \right)^2} = {\left( 3 \right)^2}
 \]
The standard equation of a circle having centre \[\left( {h,k} \right)\] and radius \[r\] is \[{\left( {x - h} \right)^2} + {\left( {x - k} \right)^2} = {r^2}\]
By comparing, we get
Thus, the centre of a circle is \[{C_2}\left( {h,k} \right) = {C_2}\left( {11,5} \right)\]
Also, the radius of a circle is \[{r_2} = 3\]
So, the distance between two centres \[{C_1}\] and \[{C_2}\] is given by
\[d\left( {{C_1},{C_2}} \right) = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} \]
Let \[{C_1}\left( {5,5} \right) = {C_1}\left( {{x_1},{y_1}} \right)\]and \[{C_2}\left( {11,5} \right) = {C_2}\left( {{x_2},{y_2}} \right)\]
Thus, we get
\[
   \Rightarrow d\left( {{C_1},{C_2}} \right) = \sqrt {{{\left( {11 - 5} \right)}^2} + {{\left( {5 - 5} \right)}^2}} \\
   \Rightarrow d\left( {{C_1},{C_2}} \right) = \sqrt {{{\left( 6 \right)}^2} + {{\left( 0 \right)}^2}} \\
   \Rightarrow d\left( {{C_1},{C_2}} \right) = \sqrt {36} \\
   \Rightarrow d\left( {{C_1},{C_2}} \right) = 6
 \]
Now, let us find the sum of the radii of two circles.
\[
  {r_1} + {r_2} = 3 + 3 \\
   \Rightarrow {r_1} + {r_2} = 6
 \]
Here, we can say that \[d\left( {{C_1},{C_2}} \right) = {r_1} + {r_2} = 6\]
Hence, these two circles have only one meeting point as they are touching externally.

Therefore, the option (C) is correct.

Additional Information: The general form of an equation of a circle is \[{x^2} + {y^2} + 2gx + 2fy + c = 0 \] with centre \[\left( { - g, - f} \right)\] and radius equal to \[{a^2} = {g^2} + {f^2} - c\].

Note: If the circles come into contact with each other on the outside, they only have one point of contact. If the circle contacts internally, they have two places of contact.