JEE MAIN - Mathematics (2019 - 9th April Morning Slot - No. 7)
All the points in the set
$$S = \left\{ {{{\alpha + i} \over {\alpha - i}}:\alpha \in R} \right\}(i = \sqrt { - 1} )$$ lie on a :
$$S = \left\{ {{{\alpha + i} \over {\alpha - i}}:\alpha \in R} \right\}(i = \sqrt { - 1} )$$ lie on a :
straight line whose slope is –1
straight line whose slope is 1.
circle whose radius is 1.
circle whose radius is $$\sqrt 2$$ .
Explanation
Let h + ik = $${{\alpha + i} \over {\alpha - i}}$$
= $${{\left( {\alpha + i} \right)\left( {\alpha + i} \right)} \over {\left( {\alpha - i} \right)\left( {\alpha + i} \right)}}$$
= $${{\left( {{\alpha ^2} - 1} \right) + 2i\alpha } \over {{\alpha ^2} + 1}}$$
$$ \therefore $$ h = $${{{\alpha ^2} - 1} \over {{\alpha ^2} + 1}}$$ and k = $${{2\alpha } \over {{\alpha ^2} + 1}}$$
By squaring and adding we get
h2 + k2 = 1
$$ \therefore $$ This is circle whose radius is 1.
= $${{\left( {\alpha + i} \right)\left( {\alpha + i} \right)} \over {\left( {\alpha - i} \right)\left( {\alpha + i} \right)}}$$
= $${{\left( {{\alpha ^2} - 1} \right) + 2i\alpha } \over {{\alpha ^2} + 1}}$$
$$ \therefore $$ h = $${{{\alpha ^2} - 1} \over {{\alpha ^2} + 1}}$$ and k = $${{2\alpha } \over {{\alpha ^2} + 1}}$$
By squaring and adding we get
h2 + k2 = 1
$$ \therefore $$ This is circle whose radius is 1.
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