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The equation Im $$\left( {{{iz - 2} \over {z - i}}} \right)$$ + 1 = 0, z $$ \in $$ C, z $$ \ne $$ i represents a part of a circle having radius equal to :

JEE · Math · previous-year question

  1. A.2
  2. B.1
  3. C.$${3 \over 4}$$correct
  4. D.$${1 \over 2}$$

Answer

C. $${3 \over 4}$$

Explanation

Let z = x + iy Then, Im $$\left( {{{iz - 2} \over {z - i}}} \right)$$ + 1 = 0 $$ \Rightarrow $$ $${\mathop{\rm Im}\nolimits} \left[ {\left( {{{i\left( {x + iy} \right) - 2} \over {x + iy - i}}} \right)} \right] + 1 = 0$$ $$ \Rightarrow $$$${\mathop{\rm Im}\nolimits} \left[ {\left( {{{ix - y - 2} \over {x + i\left( {y - 1} \right)}}} \right)} \right] + 1 = 0$$ $$ \Rightarrow $$$${\mathop{\rm Im}\nolimits} \left[ {\left( {{{ix - y - 2} \over {x + i\left( {y - 1} \right)}}} \right)\left( {{{x - i\left( {y - 1} \right)} \over {x - i\left( {y - 1} \right)}}} \right)} \right] + 1 = 0$$ $$ \Rightarrow $$$${\mathop{\rm Im}\nolimits} \left[ {\left( {{\begin{matrix} i{x^2} - {i^2}x\left( {y - 1} \right) - xy + \hfill \\ \,\,\,\,iy\left( {y - 1} \right) - 2x + i2\left( {y - 1} \right) \hfill \\\end{matrix} \over {{x^2} - {i^2}{{\left( {y - 1} \right)}^2}}}} \right)} \right] + 1 = 0$$ $$ \Rightarrow $$ $${\mathop{\rm Im}\nolimits} \left[ {\left( {{\begin{matrix} i{x^2} + x\left( {y - 1} \right) - xy \hfill \\ \,\,\, - 2x + i\left( {y - 1} \right)\left( {y + 2} \right) \hfill \\\end{matrix} \over {{x^2} + {{\left( {y - 1} \right)}^2}}}} \right)} \right] + 1 = 0$$ $$ \Rightarrow $$$${\mathop{\rm Im}\nolimits} \left[ {\left( {{\begin{matrix} x\left( {y - 1} \right) - xy\, - 2x \hfill \\ \,\, + i\left[ {\left( {y - 1} \right)\left( {y + 2} \right) + {x^2}} \right] \hfill \\\end{matrix} \over {{x^2} + {{\left( {y - 1} \right)}^2}}}} \right)} \right] + 1 = 0$$ $$ \Rightarrow $$$${{\left( {y - 1} \right)\left( {y + 2} \right) + {x^2}} \over {{x^2} + {{\left( {y - 1} \right)}^2}}} + 1 = 0$$ $$ \Rightarrow $$2x2 + 2y2 - y - 1 = 0 $$ \Rightarrow $$x2 + y2 - $$\left( {{1 \over 2}} \right)$$y - $$\left( {{1 \over 2}} \right)$$ = 0 $$ \therefore $$ Center of the circle is $$\left( {0,{1 \over 4}} \right)$$ $$ \therefore $$ Radius = $$\sqrt {{0^2} + {{\left( {{1 \over 4}} \right)}^2} + {1 \over 2}} $$ = $$\sqrt {{1 \over {16}} + {1 \over 2}} $$ = $$\sqrt {{9 \over {16}}} $$ = $${{3 \over 4}}$$

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