Question

If one end of a diameter of the circle \[\begin{array}{*{20}{c}}
  {{x^2} + {y^2} - 4x - 6y + 11}& = &0
\end{array}\]is (3, 4), then the coordinates of the opposite ends are,
A) (1, 2)
B) (2, 1)
C) (-1, 2)
D) None of these

Answer 1

Hint: Determine the center of the circle with the help of the given equation and then we will apply the formula of the midpoint of the line to evaluate the coordinates of the other end of the circle

Complete step by step solution: First of all, we will draw a figure according to the given conditions,

Figure 1
In this problem, we have to find the coordinates of the other end of the circle. Let us assume that the coordinates of the other end of the circle are C (P, Q). As shown in the figure, the diameter of the circle is line AC. Therefore, the midpoint of line AC is Point O (Center of the circle).
Now from the given equation, we will evaluate the center of the circle. Therefore,
\[\begin{array}{*{20}{c}}
  { \Rightarrow {x^2} + {y^2} - 4x - 6y + 11}& = &0
\end{array}\]
In the above equation, add and subtract 2. So, we will get
\[\begin{array}{*{20}{c}}
  { \Rightarrow {x^2} + {y^2} - 4x - 6y + 9 + 2 - 2}& = &0
\end{array}\]
\[\begin{array}{*{20}{c}}
  { \Rightarrow {x^2} + {y^2} - 4x - 6y + 13}& = &2
\end{array}\]
\[\begin{array}{*{20}{c}}
  { \Rightarrow \left( {{x^2} - 4x + 4} \right) + \left( {{y^2} - 6y + 9} \right)}& = &2
\end{array}\]
\[\begin{array}{*{20}{c}}
  { \Rightarrow {{\left( {x - 2} \right)}^2} + {{\left( {y - 3} \right)}^2}}& = &2
\end{array}\]
Now compare the above equation with the general equation of the circle.
\[\begin{array}{*{20}{c}}
  { \Rightarrow {{\left( {x - a} \right)}^2} + {{\left( {y - b} \right)}^2}}& = &{{r^2}}
\end{array}\]
We will get,
\[\begin{array}{*{20}{c}}
  { \Rightarrow a}& = &2
\end{array}\] and \[\begin{array}{*{20}{c}}
  b& = &3
\end{array}\]
Therefore, the center of the circle will be O (2, 2).
As we know, the center of the circle is the midpoint of the Line AC.
We also know the formula to determine the midpoint of the line.
\[ \Rightarrow \begin{array}{*{20}{c}}
  a& = &{\dfrac{{{x_1} + P}}{2}}
\end{array}\]And \[\begin{array}{*{20}{c}}
  b& = &{\dfrac{{{y_1} + Q}}{2}}
\end{array}\]
Therefore, the line AC
\[ \Rightarrow \begin{array}{*{20}{c}}
  2& = &{\dfrac{{3 + P}}{2}}
\end{array}\]and \[\begin{array}{*{20}{c}}
  3& = &{\dfrac{{4 + Q}}{2}}
\end{array}\]
\[ \Rightarrow \begin{array}{*{20}{c}}
  P& = &1
\end{array}\]and \[\begin{array}{*{20}{c}}
  Q& = &2
\end{array}\]
Therefore, the coordinates of the other end of the circle is (1, 2).

So, Option ‘A’ is correct

Note: It is important to note that the midpoint of the diameter of the circle always lies at the center of the circle.