Question

If $z=x+iy$, then the equation $\left|\frac{2z-i}{z+1}\right|=m$ does not represent a circle when

• A. $m=\frac{1}{2}$
• B. $m=1$
• C. $m=2$
• D. $m=3$

Answer 1

Answer: (C)

$m=2$

Let $\left|\frac{2z-i}{z+1}\right|=m$

Given, $z=x+iy$

Therefore, $\left|\frac{2(x+iy)-i}{(x+iy)+1}\right|=m$

$\Rightarrow \left|2\right(x+iy)-i|=m|x+iy)+1|$

$\Rightarrow |2x+i(2y-1\left)\right|=m\left|\right(x+1)+iy|$)

$\Rightarrow (2x{)}^2+(2y-1{)}^2=m^2\left[\right(x+1{)}^2+y^2]$

$\Rightarrow (4x^2+4y^2-4y+1=m^2(x^2+2x+1+y^2)$

$\Rightarrow 4x^2+4y^2-4y+1=m^2{x}^2+m^2{y}^2+2x+1$

On comparing we can see that for $m^2=4$ equation will be in straight line form.

So, for $m=2$ equation does not represent the circle.