[JEE Advanced 2016] let a b in mathbb r a 2 b 2 0 s z 1
Let $a,b\in\mathbb R$, $a^2+b^2\ne0$, $S=\{z=\frac1{a+ibt}:t\in\mathbb R,t\ne0\}$. If $z=x+iy\in S$, then $(x,y)$ lies on
(a) a circle of radius $\frac1{2a}$, centre $\left(\frac1{2a},0\right)$ for $a>0,b\ne0$
(b) a circle of radius $-\frac1{2a}$, centre $\left(-\frac1{2a},0\right)$ for $a<0,b\ne0$
(c) the $x$-axis for $a\ne0,b=0$
(d) the $y$-axis for $a=0,b\ne0$
1 Answer
Correct answer: (a) a circle of radius $\frac1{2a}$, centre $\left(\frac1{2a},0\right)$ for $a>0,b\ne0$
$z=\frac{a-ibt}{a^2+b^2t^2}$ traces a circle through the origin with centre $\left(\frac1{2a},0\right)$, radius $\frac1{2|a|}$ (a). If $b=0$ it is the $x$-axis (c); if $a=0$, $z$ is purely imaginary — the $y$-axis (d).
JEE Advanced 2016 · Complex Numbers — verified solution by the Vidaara Team.
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