Question

If 'p' represents $z=x+iy$ in the argand plane and $|z-1{|}^2+|z+1{|}^2=4$ then the locus of $p$ is

• A. $x^2+y^2=2$
• B. $x^2+y^2=1$
• C. $x^2+y^2=4$
• D. $x^2+y^2=3$

Answer 1

Answer: (B)

$x^2+y^2=1$

In geometric form,
${|z-1|}^2+{|z+1|}^2={|1-(-1\left)\right|}^2=4$
which is pythagoras theorem
i.e $arg\left(\frac{z-1}{z+1}\right)=\pi /2$
$\Rightarrow$ points $(1,0)$ and $(-1,0)$ are end points of diameter of the circle and $z$ lies on circumference of the circle.
Centre of circle $=(\frac{1+(-1)}{2},\frac{0+0}{2})=(0,0)$
Thus locus is,
$(x^2-0{)}^2+(y-0{)}^2=1^2$
$\Rightarrow x^2+y^2=1$