Question

If $z=\frac{3}{2+\cos \theta +i\sin \theta }$, then locus of $z$ is

• A. $x^2+y^2+4x+3=0$
• B. $x^2+y^2+4x-3=0$
• C. $x^2+y^2-4x-3=0$
• D. $x^2+y^2-4x+3=0$

Answer 1

The correct option is D $x^2+y^2-4x+3=0$

Given : $z=\frac{3}{2+\cos \theta +i\sin \theta }$
$\Rightarrow \frac{3}{z}=2+\cos \theta +i\sin \theta \Rightarrow \cos \theta +i\sin \theta =\frac{3}{z}-2$
Taking mod on both sides, we get
$\Rightarrow 1=\frac{|3-2z|}{|z|}\Rightarrow |z|=|3-2z|$
Let $z=x+iy$
$\Rightarrow |x+iy|=|(3-2x)-2iy|$
$\Rightarrow x^2+y^2=(3-2x{)}^2+4y^2\Rightarrow 3x^2+3y^2-12x+9=0\Rightarrow x^2+y^2-4x+3=0$