
If the vertices of a triangle area unit (2, −2), (−1, −1), and (5,2) then the equation of its circumcircle is?
A) \[\begin{array}{*{20}{c}}
{{x^2} + {y^2} + 3x + 3y + 8}& = &0
\end{array}\]
B) \[\begin{array}{*{20}{c}}
{{x^2} + {y^2} - 3x - 3y - 8}& = &0
\end{array}\]
C) \[\begin{array}{*{20}{c}}
{{x^2} + {y^2} - 3x + 3y + 8}& = &0
\end{array}\]
D) None of these
Answer
164.4k+ views
Hint: First determine the center of the circumcircle and then find the radius of the circumcircle. After getting all these values, apply the equation of the circle.
Formula Used: \[\begin{array}{*{20}{c}}
d& = &{\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} }
\end{array}\]
Complete step by step solution: In this problem, we have given that the vertices of the triangle are (2, −2), (−1, −1), and (5,2) respectively.
Let us assume that the center of the circumcircle is O \[\left( {{x_0},{y_0}} \right)\]and the radius of the circumcircle is R. Now, we will draw a figure according to the given data. Therefore,

Figure 1
Now according to the figure that we have drawn,
\[ \Rightarrow OA = OB = OC = R\]
To determine the center of the circumcircle, we will write,
\[ \Rightarrow \begin{array}{*{20}{c}}
{OA}& = &{OB}
\end{array}\] ………. (1)
And we know that the formula of the length of the line is,
\[ \Rightarrow \begin{array}{*{20}{c}}
d& = &{\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} }
\end{array}\]
Therefore, The length of the OA,
\[ \Rightarrow \begin{array}{*{20}{c}}
{OA}& = &{\sqrt {{{\left( {{x_0} - 2} \right)}^2} + {{\left( {{y_0} + 2} \right)}^2}} }
\end{array}\]
And OB will be,
\[ \Rightarrow \begin{array}{*{20}{c}}
{OB}& = &{\sqrt {{{\left( {{x_0} + 1} \right)}^2} + {{\left( {{y_0} + 1} \right)}^2}} }
\end{array}\]
Now from equation (1). We will get
\[ \Rightarrow \begin{array}{*{20}{c}}
{\sqrt {{{\left( {{x_0} - 2} \right)}^2} + {{\left( {{y_0} + 2} \right)}^2}} }& = &{\sqrt {{{\left( {{x_0} + 1} \right)}^2} + {{\left( {{y_0} + 1} \right)}^2}} }
\end{array}\]
Square both sides, we will get
\[ \Rightarrow \begin{array}{*{20}{c}}
{{{\left( {{x_0} - 2} \right)}^2} + {{\left( {{y_0} + 2} \right)}^2}}& = &{{{\left( {{x_0} + 1} \right)}^2} + {{\left( {{y_0} + 1} \right)}^2}}
\end{array}\]
\[\begin{array}{*{20}{c}}
{ \Rightarrow 3{x_0} - {y_0}}& = &3
\end{array}\] …………. (A).
Similarly,
For OA and OC, so we will get
\[\begin{array}{*{20}{c}}
{ \Rightarrow OA}& = &{OC}
\end{array}\]
\[\begin{array}{*{20}{c}}
{ \Rightarrow \sqrt {{{\left( {{x_0} - 2} \right)}^2} + {{\left( {{y_0} + 2} \right)}^2}} }& = &{\sqrt {{{\left( {{x_0} - 5} \right)}^2} + {{\left( {{y_0} - 2} \right)}^2}} }
\end{array}\]
Square both sides we will get
\[\begin{array}{*{20}{c}}
{ \Rightarrow {{\left( {{x_0} - 2} \right)}^2} + {{\left( {{y_0} + 2} \right)}^2}}& = &{{{\left( {{x_0} - 5} \right)}^2} + {{\left( {{y_0} - 2} \right)}^2}}
\end{array}\]
\[\begin{array}{*{20}{c}}
{ \Rightarrow 6{x_1} + 8{y_1}}& = &{21}
\end{array}\] ……….. (B)
Now from the equation (A) and (B). we will get,
\[\begin{array}{*{20}{c}}
{ \Rightarrow {x_0}}& = &{\dfrac{3}{2}}
\end{array}\] and \[\begin{array}{*{20}{c}}
{{y_0}}& = &{\dfrac{3}{2}}
\end{array}\]
Therefore, the coordinates of the center of the circumcircle are \[\left( {\dfrac{3}{2},\dfrac{3}{2}} \right)\]
Now we will determine the radius of the circumcircle. According to the figure we have,
\[ \Rightarrow OA = OB = OC = R\]
Now
\[\begin{array}{*{20}{c}}
{ \Rightarrow OA}& = &{\sqrt {{{\left( {\dfrac{3}{2} - 2} \right)}^2} + {{\left( {\dfrac{3}{2} + 2} \right)}^2}} }
\end{array}\]
\[\begin{array}{*{20}{c}}
{ \Rightarrow OA}& = &{\dfrac{5}{{\sqrt 2 }}}
\end{array}\]
Now apply the general equation of the circle, we will get
\[\begin{array}{*{20}{c}}
{ \Rightarrow {{\left( {x - {x_0}} \right)}^2} + {{\left( {y - {y_0}} \right)}^2}}& = &{{R^2}}
\end{array}\]
Now put the values,
\[\begin{array}{*{20}{c}}
{ \Rightarrow {{\left( {x - \dfrac{3}{2}} \right)}^2} + {{\left( {y - \dfrac{3}{2}} \right)}^2}}& = &{{{\left( {\dfrac{5}{{\sqrt 2 }}} \right)}^2}}
\end{array}\]
\[\begin{array}{*{20}{c}}
{ \Rightarrow {x^2} + {y^2} - 3x - 3y - 8}& = &0
\end{array}\]
So, Option ‘B’ is correct
Note: It is important to note that the length of the lines OA, OB and OC will be equal (radii of the circle).
Formula Used: \[\begin{array}{*{20}{c}}
d& = &{\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} }
\end{array}\]
Complete step by step solution: In this problem, we have given that the vertices of the triangle are (2, −2), (−1, −1), and (5,2) respectively.
Let us assume that the center of the circumcircle is O \[\left( {{x_0},{y_0}} \right)\]and the radius of the circumcircle is R. Now, we will draw a figure according to the given data. Therefore,

Figure 1
Now according to the figure that we have drawn,
\[ \Rightarrow OA = OB = OC = R\]
To determine the center of the circumcircle, we will write,
\[ \Rightarrow \begin{array}{*{20}{c}}
{OA}& = &{OB}
\end{array}\] ………. (1)
And we know that the formula of the length of the line is,
\[ \Rightarrow \begin{array}{*{20}{c}}
d& = &{\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} }
\end{array}\]
Therefore, The length of the OA,
\[ \Rightarrow \begin{array}{*{20}{c}}
{OA}& = &{\sqrt {{{\left( {{x_0} - 2} \right)}^2} + {{\left( {{y_0} + 2} \right)}^2}} }
\end{array}\]
And OB will be,
\[ \Rightarrow \begin{array}{*{20}{c}}
{OB}& = &{\sqrt {{{\left( {{x_0} + 1} \right)}^2} + {{\left( {{y_0} + 1} \right)}^2}} }
\end{array}\]
Now from equation (1). We will get
\[ \Rightarrow \begin{array}{*{20}{c}}
{\sqrt {{{\left( {{x_0} - 2} \right)}^2} + {{\left( {{y_0} + 2} \right)}^2}} }& = &{\sqrt {{{\left( {{x_0} + 1} \right)}^2} + {{\left( {{y_0} + 1} \right)}^2}} }
\end{array}\]
Square both sides, we will get
\[ \Rightarrow \begin{array}{*{20}{c}}
{{{\left( {{x_0} - 2} \right)}^2} + {{\left( {{y_0} + 2} \right)}^2}}& = &{{{\left( {{x_0} + 1} \right)}^2} + {{\left( {{y_0} + 1} \right)}^2}}
\end{array}\]
\[\begin{array}{*{20}{c}}
{ \Rightarrow 3{x_0} - {y_0}}& = &3
\end{array}\] …………. (A).
Similarly,
For OA and OC, so we will get
\[\begin{array}{*{20}{c}}
{ \Rightarrow OA}& = &{OC}
\end{array}\]
\[\begin{array}{*{20}{c}}
{ \Rightarrow \sqrt {{{\left( {{x_0} - 2} \right)}^2} + {{\left( {{y_0} + 2} \right)}^2}} }& = &{\sqrt {{{\left( {{x_0} - 5} \right)}^2} + {{\left( {{y_0} - 2} \right)}^2}} }
\end{array}\]
Square both sides we will get
\[\begin{array}{*{20}{c}}
{ \Rightarrow {{\left( {{x_0} - 2} \right)}^2} + {{\left( {{y_0} + 2} \right)}^2}}& = &{{{\left( {{x_0} - 5} \right)}^2} + {{\left( {{y_0} - 2} \right)}^2}}
\end{array}\]
\[\begin{array}{*{20}{c}}
{ \Rightarrow 6{x_1} + 8{y_1}}& = &{21}
\end{array}\] ……….. (B)
Now from the equation (A) and (B). we will get,
\[\begin{array}{*{20}{c}}
{ \Rightarrow {x_0}}& = &{\dfrac{3}{2}}
\end{array}\] and \[\begin{array}{*{20}{c}}
{{y_0}}& = &{\dfrac{3}{2}}
\end{array}\]
Therefore, the coordinates of the center of the circumcircle are \[\left( {\dfrac{3}{2},\dfrac{3}{2}} \right)\]
Now we will determine the radius of the circumcircle. According to the figure we have,
\[ \Rightarrow OA = OB = OC = R\]
Now
\[\begin{array}{*{20}{c}}
{ \Rightarrow OA}& = &{\sqrt {{{\left( {\dfrac{3}{2} - 2} \right)}^2} + {{\left( {\dfrac{3}{2} + 2} \right)}^2}} }
\end{array}\]
\[\begin{array}{*{20}{c}}
{ \Rightarrow OA}& = &{\dfrac{5}{{\sqrt 2 }}}
\end{array}\]
Now apply the general equation of the circle, we will get
\[\begin{array}{*{20}{c}}
{ \Rightarrow {{\left( {x - {x_0}} \right)}^2} + {{\left( {y - {y_0}} \right)}^2}}& = &{{R^2}}
\end{array}\]
Now put the values,
\[\begin{array}{*{20}{c}}
{ \Rightarrow {{\left( {x - \dfrac{3}{2}} \right)}^2} + {{\left( {y - \dfrac{3}{2}} \right)}^2}}& = &{{{\left( {\dfrac{5}{{\sqrt 2 }}} \right)}^2}}
\end{array}\]
\[\begin{array}{*{20}{c}}
{ \Rightarrow {x^2} + {y^2} - 3x - 3y - 8}& = &0
\end{array}\]
So, Option ‘B’ is correct
Note: It is important to note that the length of the lines OA, OB and OC will be equal (radii 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

What is Normality in Chemistry?

Chemistry Electronic Configuration of D Block Elements: JEE Main 2025

Other Pages
NCERT Solutions for Class 11 Maths Chapter 6 Permutations and Combinations

NCERT Solutions for Class 11 Maths Chapter 8 Sequences and Series

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
