
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
164.1k+ views
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.
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.
Recently Updated Pages
Geometry of Complex Numbers – Topics, Reception, Audience and Related Readings

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Main 2025 Session 2: Exam Date, Admit Card, Syllabus, & More

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

Trending doubts
Degree of Dissociation and Its Formula With Solved Example for JEE

Instantaneous Velocity - Formula based Examples for JEE

JEE Main Chemistry Question Paper with Answer Keys and Solutions

JEE Main Reservation Criteria 2025: SC, ST, EWS, and PwD Candidates

JEE Mains 2025 Cut-Off GFIT: Check All Rounds Cutoff Ranks

Lami's Theorem

Other Pages
Total MBBS Seats in India 2025: Government College Seat Matrix

NEET Total Marks 2025: Important Information and Key Updates

Neet Cut Off 2025 for MBBS in Tamilnadu: AIQ & State Quota Analysis

Karnataka NEET Cut off 2025 - Category Wise Cut Off Marks

NEET Marks vs Rank 2024|How to Calculate?

NEET 2025: All Major Changes in Application Process, Pattern and More
