
Let complex numbers $\alpha {\text{ and }}\dfrac{1}{\alpha }$ lie on circles ${\left( {x - {x_o}} \right)^2} + {\left( {y - {y_o}} \right)^2} = {r^2}$ and ${\left( {x - {x_o}} \right)^2} + {\left( {y - {y_o}} \right)^2} = 4{r^2}$, respectively. If ${z_o} = {x_o} + i{y_o}$ satisfies the equation, $2{\left| {{z_o}} \right|^2} = {r^2} + 2$, then $\left| \alpha \right|$ =
$\left( a \right)\dfrac{1}{{\sqrt 2 }}$
$\left( b \right)\dfrac{1}{2}$
$\left( c \right)\dfrac{1}{{\sqrt 7 }}$
$\left( d \right)\dfrac{1}{3}$
Answer
134.1k+ views
Hint: In this particular question use the concept that in the standard equation of the circle ${\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}$, (h, k) and r are the center and the radius of the circle respectively and in complex plane we can write the equation of the circle, $\left| {z - {z_o}} \right| = r$ where z and ${z_o}$ both are the complex numbers and r is the radius of the circle, so use these concepts to reach the solution of the question.
Complete step-by-step answer:
Given data:
Complex numbers $\alpha {\text{ and }}\dfrac{1}{\alpha }$ lie on circles ${\left( {x - {x_o}} \right)^2} + {\left( {y - {y_o}} \right)^2} = {r^2}$ and ${\left( {x - {x_o}} \right)^2} + {\left( {y - {y_o}} \right)^2} = 4{r^2}$.
So the equations of the circles in the complex plane is written as,
$ \Rightarrow \left| {z - {z_o}} \right| = r$ and $\left| {z - {z_o}} \right| = 2r$
Now $\alpha {\text{ and }}\dfrac{1}{\alpha }$ which represents the complex number and lie on the circle so it satisfies the above equation respectively, so we have,
$ \Rightarrow \left| {\alpha - {z_o}} \right| = r$, and $\left| {\dfrac{1}{\alpha } - {z_o}} \right| = 2r$
$ \Rightarrow \left| {\alpha - {z_o}} \right| = r$, and $\dfrac{{\left| {1 - \alpha {z_o}} \right|}}{{\left| \alpha \right|}} = 2r$
$ \Rightarrow \left| {\alpha - {z_o}} \right| = r$, and $\left| {1 - \alpha {z_o}} \right| = 2r\left| \alpha \right|$
Now squaring on both sides in both of the above equation we have,
$ \Rightarrow {\left| {\alpha - {z_o}} \right|^2} = {r^2}$, and ${\left| {1 - \alpha {z_o}} \right|^2} = 4{r^2}{\left| \alpha \right|^2}$
Now as we know that ${\left| {a - b} \right|^2} = \left( {a - b} \right)\left( {\overline {a - b} } \right) = \left( {a - b} \right)\left( {\bar a - \bar b} \right)$ and ${\left| {1 - ab} \right|^2} = \left( {1 - \bar ab} \right)\left( {1 - a\bar b} \right)$ where a and b both represents the complex numbers so use these properties in the above equation we have,
$ \Rightarrow \left( {\alpha - {z_o}} \right)\left( {\overline {\alpha - {z_o}} } \right) = {r^2}$ , and $\left( {1 - \bar \alpha {z_o}} \right)\left( {1 - \alpha {{\bar z}_o}} \right) = 4{r^2}{\left| \alpha \right|^2}$
\[ \Rightarrow \left( {\alpha - {z_o}} \right)\left( {\bar \alpha - {{\bar z}_o}} \right) = {r^2}\] , and $\left( {1 - \bar \alpha {z_o}} \right)\left( {1 - \alpha {{\bar z}_o}} \right) = 4{r^2}{\left| \alpha \right|^2}$
Now simplify it we have,
\[ \Rightarrow \left( {\alpha - {z_o}} \right)\left( {\bar \alpha - {{\bar z}_o}} \right) = {r^2}\]
\[ \Rightarrow \alpha \bar \alpha - \alpha {\bar z_o} - {z_o}\bar \alpha + {z_o}{\bar z_o} = {r^2}\]
\[ \Rightarrow \alpha {\bar z_o} + {z_o}\bar \alpha = \alpha \bar \alpha + {z_o}{\bar z_o} - {r^2}\]
\[ \Rightarrow \alpha {\bar z_o} + {z_o}\bar \alpha = {\left| \alpha \right|^2} + {\left| {{z_o}} \right|^2} - {r^2}\]..................... (1), \[\left[ {\because \alpha \bar \alpha = {{\left| \alpha \right|}^2},{z_o}{{\bar z}_o} = {{\left| {{z_o}} \right|}^2}} \right]\]
And
$ \Rightarrow \left( {1 - \bar \alpha {z_o}} \right)\left( {1 - \alpha {{\bar z}_o}} \right) = 4{r^2}{\left| \alpha \right|^2}$
$ \Rightarrow 1 - \alpha {\bar z_o} - {z_o}\bar \alpha + \alpha \bar \alpha {z_o}{\bar z_o} = 4{r^2}{\left| \alpha \right|^2}$
$ \Rightarrow 1 - \alpha {\bar z_o} - {z_o}\bar \alpha + {\left| \alpha \right|^2}{\left| {{z_o}} \right|^2} = 4{r^2}{\left| \alpha \right|^2}$
$ \Rightarrow \alpha {\bar z_o} + {z_o}\bar \alpha = 1 + {\left| \alpha \right|^2}{\left| {{z_o}} \right|^2} - 4{r^2}{\left| \alpha \right|^2}$................... (2)
Now as we see that the LHS of the equation (1) and (2) are same so equate RHS of the equation (1) and (2) we have,
\[ \Rightarrow {\left| \alpha \right|^2} + {\left| {{z_o}} \right|^2} - {r^2} = 1 + {\left| \alpha \right|^2}{\left| {{z_o}} \right|^2} - 4{r^2}{\left| \alpha \right|^2}\].............. (3)
Now it is given that $2{\left| {{z_o}} \right|^2} = {r^2} + 2$, $ \Rightarrow {\left| {{z_o}} \right|^2} = \dfrac{{{r^2} + 2}}{2}$ so substitute this value in equation (3) we have,
\[ \Rightarrow {\left| \alpha \right|^2} + \dfrac{{{r^2} + 2}}{2} - {r^2} = 1 + {\left| \alpha \right|^2}\left( {\dfrac{{{r^2} + 2}}{2}} \right) - 4{r^2}{\left| \alpha \right|^2}\]
Now simplify it we have,
\[ \Rightarrow {\left| \alpha \right|^2} - \dfrac{{{r^2}}}{2} + 1 = 1 + \dfrac{{{{\left| \alpha \right|}^2}{r^2}}}{2} + {\left| \alpha \right|^2} - 4{r^2}{\left| \alpha \right|^2}\]
Now cancel out the same terms with same sign from LHS and RHS we have,
\[ \Rightarrow - \dfrac{{{r^2}}}{2} = \dfrac{{{{\left| \alpha \right|}^2}{r^2}}}{2} - 4{r^2}{\left| \alpha \right|^2}\]
Now divide by ${r^2}$ throughout we have,
\[ \Rightarrow - \dfrac{1}{2} = \dfrac{{{{\left| \alpha \right|}^2}}}{2} - 4{\left| \alpha \right|^2}\]
\[ \Rightarrow - \dfrac{1}{2} = - \dfrac{{7{{\left| \alpha \right|}^2}}}{2}\]
\[ \Rightarrow 7{\left| \alpha \right|^2} = 1\]
\[ \Rightarrow {\left| \alpha \right|^2} = \dfrac{1}{7}\]
Now take square root on both sides we have,
\[ \Rightarrow \left| \alpha \right| = \sqrt {\dfrac{1}{7}} = \dfrac{1}{{\sqrt 7 }}\]
So this is the required answer.
Hence option (c) is the correct answer.
Note: Whenever we face such types of questions the key concept we have to remember is that always recall that ${\left| {a - b} \right|^2} = \left( {a - b} \right)\left( {\overline {a - b} } \right) = \left( {a - b} \right)\left( {\bar a - \bar b} \right)$ and ${\left| {1 - ab} \right|^2} = \left( {1 - \bar ab} \right)\left( {1 - a\bar b} \right)$ where a and b both represents the complex numbers, so according to this property simplify the equation as above and then equate the RHS of the above equations as LHS are same then again simplify the equation using the given condition in the question as above we will get the required answer.
Complete step-by-step answer:
Given data:
Complex numbers $\alpha {\text{ and }}\dfrac{1}{\alpha }$ lie on circles ${\left( {x - {x_o}} \right)^2} + {\left( {y - {y_o}} \right)^2} = {r^2}$ and ${\left( {x - {x_o}} \right)^2} + {\left( {y - {y_o}} \right)^2} = 4{r^2}$.
So the equations of the circles in the complex plane is written as,
$ \Rightarrow \left| {z - {z_o}} \right| = r$ and $\left| {z - {z_o}} \right| = 2r$
Now $\alpha {\text{ and }}\dfrac{1}{\alpha }$ which represents the complex number and lie on the circle so it satisfies the above equation respectively, so we have,
$ \Rightarrow \left| {\alpha - {z_o}} \right| = r$, and $\left| {\dfrac{1}{\alpha } - {z_o}} \right| = 2r$
$ \Rightarrow \left| {\alpha - {z_o}} \right| = r$, and $\dfrac{{\left| {1 - \alpha {z_o}} \right|}}{{\left| \alpha \right|}} = 2r$
$ \Rightarrow \left| {\alpha - {z_o}} \right| = r$, and $\left| {1 - \alpha {z_o}} \right| = 2r\left| \alpha \right|$
Now squaring on both sides in both of the above equation we have,
$ \Rightarrow {\left| {\alpha - {z_o}} \right|^2} = {r^2}$, and ${\left| {1 - \alpha {z_o}} \right|^2} = 4{r^2}{\left| \alpha \right|^2}$
Now as we know that ${\left| {a - b} \right|^2} = \left( {a - b} \right)\left( {\overline {a - b} } \right) = \left( {a - b} \right)\left( {\bar a - \bar b} \right)$ and ${\left| {1 - ab} \right|^2} = \left( {1 - \bar ab} \right)\left( {1 - a\bar b} \right)$ where a and b both represents the complex numbers so use these properties in the above equation we have,
$ \Rightarrow \left( {\alpha - {z_o}} \right)\left( {\overline {\alpha - {z_o}} } \right) = {r^2}$ , and $\left( {1 - \bar \alpha {z_o}} \right)\left( {1 - \alpha {{\bar z}_o}} \right) = 4{r^2}{\left| \alpha \right|^2}$
\[ \Rightarrow \left( {\alpha - {z_o}} \right)\left( {\bar \alpha - {{\bar z}_o}} \right) = {r^2}\] , and $\left( {1 - \bar \alpha {z_o}} \right)\left( {1 - \alpha {{\bar z}_o}} \right) = 4{r^2}{\left| \alpha \right|^2}$
Now simplify it we have,
\[ \Rightarrow \left( {\alpha - {z_o}} \right)\left( {\bar \alpha - {{\bar z}_o}} \right) = {r^2}\]
\[ \Rightarrow \alpha \bar \alpha - \alpha {\bar z_o} - {z_o}\bar \alpha + {z_o}{\bar z_o} = {r^2}\]
\[ \Rightarrow \alpha {\bar z_o} + {z_o}\bar \alpha = \alpha \bar \alpha + {z_o}{\bar z_o} - {r^2}\]
\[ \Rightarrow \alpha {\bar z_o} + {z_o}\bar \alpha = {\left| \alpha \right|^2} + {\left| {{z_o}} \right|^2} - {r^2}\]..................... (1), \[\left[ {\because \alpha \bar \alpha = {{\left| \alpha \right|}^2},{z_o}{{\bar z}_o} = {{\left| {{z_o}} \right|}^2}} \right]\]
And
$ \Rightarrow \left( {1 - \bar \alpha {z_o}} \right)\left( {1 - \alpha {{\bar z}_o}} \right) = 4{r^2}{\left| \alpha \right|^2}$
$ \Rightarrow 1 - \alpha {\bar z_o} - {z_o}\bar \alpha + \alpha \bar \alpha {z_o}{\bar z_o} = 4{r^2}{\left| \alpha \right|^2}$
$ \Rightarrow 1 - \alpha {\bar z_o} - {z_o}\bar \alpha + {\left| \alpha \right|^2}{\left| {{z_o}} \right|^2} = 4{r^2}{\left| \alpha \right|^2}$
$ \Rightarrow \alpha {\bar z_o} + {z_o}\bar \alpha = 1 + {\left| \alpha \right|^2}{\left| {{z_o}} \right|^2} - 4{r^2}{\left| \alpha \right|^2}$................... (2)
Now as we see that the LHS of the equation (1) and (2) are same so equate RHS of the equation (1) and (2) we have,
\[ \Rightarrow {\left| \alpha \right|^2} + {\left| {{z_o}} \right|^2} - {r^2} = 1 + {\left| \alpha \right|^2}{\left| {{z_o}} \right|^2} - 4{r^2}{\left| \alpha \right|^2}\].............. (3)
Now it is given that $2{\left| {{z_o}} \right|^2} = {r^2} + 2$, $ \Rightarrow {\left| {{z_o}} \right|^2} = \dfrac{{{r^2} + 2}}{2}$ so substitute this value in equation (3) we have,
\[ \Rightarrow {\left| \alpha \right|^2} + \dfrac{{{r^2} + 2}}{2} - {r^2} = 1 + {\left| \alpha \right|^2}\left( {\dfrac{{{r^2} + 2}}{2}} \right) - 4{r^2}{\left| \alpha \right|^2}\]
Now simplify it we have,
\[ \Rightarrow {\left| \alpha \right|^2} - \dfrac{{{r^2}}}{2} + 1 = 1 + \dfrac{{{{\left| \alpha \right|}^2}{r^2}}}{2} + {\left| \alpha \right|^2} - 4{r^2}{\left| \alpha \right|^2}\]
Now cancel out the same terms with same sign from LHS and RHS we have,
\[ \Rightarrow - \dfrac{{{r^2}}}{2} = \dfrac{{{{\left| \alpha \right|}^2}{r^2}}}{2} - 4{r^2}{\left| \alpha \right|^2}\]
Now divide by ${r^2}$ throughout we have,
\[ \Rightarrow - \dfrac{1}{2} = \dfrac{{{{\left| \alpha \right|}^2}}}{2} - 4{\left| \alpha \right|^2}\]
\[ \Rightarrow - \dfrac{1}{2} = - \dfrac{{7{{\left| \alpha \right|}^2}}}{2}\]
\[ \Rightarrow 7{\left| \alpha \right|^2} = 1\]
\[ \Rightarrow {\left| \alpha \right|^2} = \dfrac{1}{7}\]
Now take square root on both sides we have,
\[ \Rightarrow \left| \alpha \right| = \sqrt {\dfrac{1}{7}} = \dfrac{1}{{\sqrt 7 }}\]
So this is the required answer.
Hence option (c) is the correct answer.
Note: Whenever we face such types of questions the key concept we have to remember is that always recall that ${\left| {a - b} \right|^2} = \left( {a - b} \right)\left( {\overline {a - b} } \right) = \left( {a - b} \right)\left( {\bar a - \bar b} \right)$ and ${\left| {1 - ab} \right|^2} = \left( {1 - \bar ab} \right)\left( {1 - a\bar b} \right)$ where a and b both represents the complex numbers, so according to this property simplify the equation as above and then equate the RHS of the above equations as LHS are same then again simplify the equation using the given condition in the question as above we will get the required answer.
Recently Updated Pages
JEE Main 2025 Session 2 Form Correction (Closed) – What Can Be Edited

Sign up for JEE Main 2025 Live Classes - Vedantu

JEE Main Books 2023-24: Best JEE Main Books for Physics, Chemistry and Maths

JEE Main 2023 April 13 Shift 1 Question Paper with Answer Key

JEE Main 2023 April 11 Shift 2 Question Paper with Answer Key

JEE Main 2023 April 10 Shift 2 Question Paper with Answer Key

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

The area of an expanding rectangle is increasing at class 12 maths JEE_Main

JEE Main Chemistry Question Paper with Answer Keys and Solutions

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced 2024 Syllabus Weightage

Free Radical Substitution Mechanism of Alkanes for JEE Main 2025

Current Loop as Magnetic Dipole and Its Derivation for JEE

Inertial and Non-Inertial Frame of Reference - JEE Important Topic

CBSE Class 12 English Core Syllabus 2024-25: Updated Curriculum
