Question

Let $a,b \in R$ and ${a^2} + {b^2} \ne 0$. Suppose $S\left\{ {z \in C:z = \dfrac{1}{{a + ibt}},t \in R,t \ne 0} \right\},$ where $i = \sqrt { - 1} $. If $z = x + iy$ and $z \in S$, then $(x,y)$ lies on
A) The circle with radius $ \dfrac{1}{{2a}}{\text{ and centre }}\left( {\dfrac{1}{{2a}},0} \right){\text{ for }}a > 0,b \ne 0 \\$
B) The circle with radius $\dfrac{1}{{2a}}{\text{ and centre }}\left( { - \dfrac{1}{{2a}},0} \right){\text{ for }}a < 0,b \ne 0 \\$
C) The x - axis for $a \ne 0,b = 0 \\$
D) The y - axis for $a = 0,b \ne 0 \\$

Answer 1

Hint: This problem is related to Complex numbers. Complex numbers are represented as$a \pm ib$, where ‘$a$’ is the real part of the complex number and ‘$ib$’ is the complex part of the complex number. To simplify the complex numbers, try to solve and eliminate the complex part of the equations. Then, compare this equation with general equation of circle with centre at $(h,k)$and radius$r$which is ${(x - h)^2} + {(y - k)^2} = {r^2}$ to get the desired result.

Complete step by step answer:
From the problem, there are two complex number equations given. We write the expression given in the question as,
$z = \dfrac{1}{{a + ibt}}$ and $x + iy$, thus the equation can be written as,
$x + iy = \dfrac{1}{{a + ibt}}$
Now, multiply and divide the RHS of the above equation by $a - ibt$, we get

$x + iy = \dfrac{1}{{a + ibt}} \times \dfrac{{a - ibt}}{{a - ibt}}$
Simplifying it by using formula $(a + b)(a - b) = {a^2} - {b^2}$ in the denominator, we get
$
  x + iy = \dfrac{{a - ibt}}{{{a^2} - {{(ibt)}^2}}} \\
  x + iy = \dfrac{{a - ibt}}{{{a^2} + {{(bt)}^2}}}{\text{ (using the property, }}i = \sqrt { - 1} {\text{ and }}{i^2} = - 1) \\
  x + iy = \dfrac{{a - ibt}}{{{a^2} + {b^2}{t^2}}} \\
 $
Now, separating the real and complex numbers from the above expression, we get
$
  x = \dfrac{a}{{{a^2} + {b^2}{t^2}}}{\text{ }}..........................{\text{ (1)}} \\
  y = - \dfrac{{bt}}{{{a^2} + {b^2}{t^2}}}{\text{ }}.........................{\text{ (2)}} \\
 $
Now, if $a = 0,b \ne 0$ then from the eq. (1) we get $x = 0$, which means the point $(x,y)$ lies on the y-axis.

Hence, option (D) is correct.

Now, if $a \ne 0,b = 0$ then from the eq. (2) we get $y = 0$, which means the point $(x,y)$ lies on the x-axis.

Hence, option (C) is correct.

Now, from eq. (1), after rearranging we get the following expression,
${a^2} + {b^2}{t^2} = \dfrac{a}{x}$, putting this value in eq. (2), we get
$
  y = - \dfrac{{bt}}{{{\raise0.7ex\hbox{$a$}
\!\lower0.7ex\hbox{$x$}}}} \\
  y\dfrac{a}{x} = - bt \\
  t = - \dfrac{{ay}}{{bx}} \\
 $
Now, putting the value of $t$ in eq. (1) and rearranging we get
$
  x = \dfrac{a}{{{a^2} + {b^2}{{( - \dfrac{{ay}}{{bx}})}^2}}}{\text{ }} \\
  x\left( {{a^2} + {b^2}{{( - \dfrac{{ay}}{{bx}})}^2}} \right) = a \\
  x\left( {{a^2}{b^2}{x^2} + {b^2}{a^2}{y^2}} \right) = {b^2}{x^2}a \\
  {a^2}{b^2}{x^2} + {b^2}{a^2}{y^2} = {b^2}xa{\text{ }}.............{\text{ (3)}} \\
 $
Dividing the above eq. (3) by ${b^2}{a^2}$ we get
$
  {x^2} + {y^2} = \dfrac{x}{a} \\
   \Rightarrow {x^2} + {y^2} - \dfrac{x}{a} = 0 \\
 $
Now, by adding and subtracting $\dfrac{1}{{4{a^2}}}$ in the above equation and rearranging in the following way, we get,
$
   \Rightarrow {x^2} - \dfrac{x}{a} + \dfrac{1}{{4{a^2}}} - \dfrac{1}{{4{a^2}}} + {y^2} = 0 \\
   \Rightarrow {(x - \dfrac{1}{{2a}})^2} + {(y - 0)^2} = \dfrac{1}{{4{a^2}}} \\
   \Rightarrow {(x - \dfrac{1}{{2a}})^2} + {(y - 0)^2} = {\left( {\dfrac{1}{{2a}}} \right)^2} \\
 $
 We can compare this equation with general equation of circle with centre at$(h,k)$and radius$r$which is
${(x - h)^2} + {(y - k)^2} = {r^2}$, we get
Centre of the circle$\left( {\dfrac{1}{{2a}},0} \right){\text{ and radius }}\dfrac{1}{{2a}}$

This way option (A) is correct.

So the correct options to the problem are (A), (C) and (D).

Note: It is a very common practice to solve these types of problems with the help of standard formulae for complex numbers and circles. You shall have to identify at which place these formulae can be used in particular problems.