JEE MAIN - Mathematics (2017 - 9th April Morning Slot - No. 5)
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 :
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 :
2
1
$${3 \over 4}$$
$${1 \over 2}$$
Explanation
Let z = x + iy
Then,
Im $$\left( {{{iz - 2} \over {z - i}}} \right)$$ + 1 = 0
$$ \Rightarrow $$ $${\mathop{\rm Im}\nolimits} \left[ {\left( {{{i\left( {x + iy} \right) - 2} \over {x + iy - i}}} \right)} \right] + 1 = 0$$
$$ \Rightarrow $$$${\mathop{\rm Im}\nolimits} \left[ {\left( {{{ix - y - 2} \over {x + i\left( {y - 1} \right)}}} \right)} \right] + 1 = 0$$
$$ \Rightarrow $$$${\mathop{\rm Im}\nolimits} \left[ {\left( {{{ix - y - 2} \over {x + i\left( {y - 1} \right)}}} \right)\left( {{{x - i\left( {y - 1} \right)} \over {x - i\left( {y - 1} \right)}}} \right)} \right] + 1 = 0$$
$$ \Rightarrow $$$${\mathop{\rm Im}\nolimits} \left[ {\left( {{\matrix{ i{x^2} - {i^2}x\left( {y - 1} \right) - xy + \hfill \cr \,\,\,\,iy\left( {y - 1} \right) - 2x + i2\left( {y - 1} \right) \hfill \cr} \over {{x^2} - {i^2}{{\left( {y - 1} \right)}^2}}}} \right)} \right] + 1 = 0$$
$$ \Rightarrow $$ $${\mathop{\rm Im}\nolimits} \left[ {\left( {{\matrix{ i{x^2} + x\left( {y - 1} \right) - xy \hfill \cr \,\,\, - 2x + i\left( {y - 1} \right)\left( {y + 2} \right) \hfill \cr} \over {{x^2} + {{\left( {y - 1} \right)}^2}}}} \right)} \right] + 1 = 0$$
$$ \Rightarrow $$$${\mathop{\rm Im}\nolimits} \left[ {\left( {{\matrix{ x\left( {y - 1} \right) - xy\, - 2x \hfill \cr \,\, + i\left[ {\left( {y - 1} \right)\left( {y + 2} \right) + {x^2}} \right] \hfill \cr} \over {{x^2} + {{\left( {y - 1} \right)}^2}}}} \right)} \right] + 1 = 0$$
$$ \Rightarrow $$$${{\left( {y - 1} \right)\left( {y + 2} \right) + {x^2}} \over {{x^2} + {{\left( {y - 1} \right)}^2}}} + 1 = 0$$
$$ \Rightarrow $$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 = $$\sqrt {{0^2} + {{\left( {{1 \over 4}} \right)}^2} + {1 \over 2}} $$
= $$\sqrt {{1 \over {16}} + {1 \over 2}} $$
= $$\sqrt {{9 \over {16}}} $$
= $${{3 \over 4}}$$
Then,
Im $$\left( {{{iz - 2} \over {z - i}}} \right)$$ + 1 = 0
$$ \Rightarrow $$ $${\mathop{\rm Im}\nolimits} \left[ {\left( {{{i\left( {x + iy} \right) - 2} \over {x + iy - i}}} \right)} \right] + 1 = 0$$
$$ \Rightarrow $$$${\mathop{\rm Im}\nolimits} \left[ {\left( {{{ix - y - 2} \over {x + i\left( {y - 1} \right)}}} \right)} \right] + 1 = 0$$
$$ \Rightarrow $$$${\mathop{\rm Im}\nolimits} \left[ {\left( {{{ix - y - 2} \over {x + i\left( {y - 1} \right)}}} \right)\left( {{{x - i\left( {y - 1} \right)} \over {x - i\left( {y - 1} \right)}}} \right)} \right] + 1 = 0$$
$$ \Rightarrow $$$${\mathop{\rm Im}\nolimits} \left[ {\left( {{\matrix{ i{x^2} - {i^2}x\left( {y - 1} \right) - xy + \hfill \cr \,\,\,\,iy\left( {y - 1} \right) - 2x + i2\left( {y - 1} \right) \hfill \cr} \over {{x^2} - {i^2}{{\left( {y - 1} \right)}^2}}}} \right)} \right] + 1 = 0$$
$$ \Rightarrow $$ $${\mathop{\rm Im}\nolimits} \left[ {\left( {{\matrix{ i{x^2} + x\left( {y - 1} \right) - xy \hfill \cr \,\,\, - 2x + i\left( {y - 1} \right)\left( {y + 2} \right) \hfill \cr} \over {{x^2} + {{\left( {y - 1} \right)}^2}}}} \right)} \right] + 1 = 0$$
$$ \Rightarrow $$$${\mathop{\rm Im}\nolimits} \left[ {\left( {{\matrix{ x\left( {y - 1} \right) - xy\, - 2x \hfill \cr \,\, + i\left[ {\left( {y - 1} \right)\left( {y + 2} \right) + {x^2}} \right] \hfill \cr} \over {{x^2} + {{\left( {y - 1} \right)}^2}}}} \right)} \right] + 1 = 0$$
$$ \Rightarrow $$$${{\left( {y - 1} \right)\left( {y + 2} \right) + {x^2}} \over {{x^2} + {{\left( {y - 1} \right)}^2}}} + 1 = 0$$
$$ \Rightarrow $$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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