JEE MAIN - Mathematics (2021 - 18th March Morning Shift - No. 8)
If the equation $$a|z{|^2} + \overline {\overline \alpha z + \alpha \overline z } + d = 0$$ represents a circle where a, d are real constants then which of the following condition is correct?
|$$\alpha$$|2 $$-$$ ad $$\ne$$ 0
|$$\alpha$$|2 $$-$$ ad > 0 and a$$\in$$R $$-$$ {0}
|$$\alpha$$|2 $$-$$ ad $$ \ge $$ 0 and a$$\in$$R
$$\alpha$$ = 0, a, d$$\in$$R+
Explanation
$$a|z{|^2} + \alpha \overline z + \overline \alpha z + d = 0$$
$$ \Rightarrow $$ $$z\overline z + \left( {{\alpha \over a}} \right)\overline z + \left( {{{\overline \alpha } \over a}} \right)z + {d \over a} = 0$$
$$ \therefore $$ Centre $$ = - {\alpha \over a}$$
$$r = \sqrt {{{\left| {{\alpha \over a}} \right|}^2} - {d \over a}} $$
$$ \Rightarrow {\left| {{\alpha \over a}} \right|^2} \ge {d \over a}$$
$$ \Rightarrow {\left| \alpha \right|^2} \ge ad$$
$$ \Rightarrow $$ $$z\overline z + \left( {{\alpha \over a}} \right)\overline z + \left( {{{\overline \alpha } \over a}} \right)z + {d \over a} = 0$$
$$ \therefore $$ Centre $$ = - {\alpha \over a}$$
$$r = \sqrt {{{\left| {{\alpha \over a}} \right|}^2} - {d \over a}} $$
$$ \Rightarrow {\left| {{\alpha \over a}} \right|^2} \ge {d \over a}$$
$$ \Rightarrow {\left| \alpha \right|^2} \ge ad$$
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