Question

If $\frac{z-4}{z-2i}$ is purely imaginary then the locus of $z$ is

• A. $(x-2{)}^2+(y-1{)}^2=5$
• B. $(x-2{)}^2+(y+1{)}^2=5$
• C. $(x+2{)}^2+(y-1{)}^2=5$
• D. $(x+2{)}^2+(y+1{)}^2=5$

Answer 1

The correct option is A $(x-2{)}^2+(y-1{)}^2=5$

$\frac{z-4}{z-2i}$ is purely imaginary
$\Rightarrow \frac{z-4}{z-2i}+\overline{\frac{z-4}{z-2i}}=0[m+\overline{m}=2Re(m)=0]$
$\Rightarrow (z-4)(\overline{z}+2i)+(\overline{z}-4)(z-2i)=0$
$\Rightarrow |z{|}^2+2iz-4\overline{z}-8i+|z{|}^2-2i\overline{z}-4z+8i=0$

Taking $z=x+iy$
$\Rightarrow 2(x^2+y^2)+\left\{2i\right(x+iy)-2i(x-iy\left)\right\}-4(x-iy+x+iy)=0$
$\Rightarrow 2(x^2+y^2)+4i^{2}y-8x=0$

$\Rightarrow x^2+y^2-2y-4x=0$
$\Rightarrow (x-2{)}^2+(y-1{)}^2=5$