Question

The equation $Im\left(\frac{iz-2}{z-i}\right)+1=0,zϵC,z\ne i$ represents a part of a circle having radius equal to.

• A. $\frac{1}{2}$
• B. $1$
• C. $2$
• D. $\frac{3}{4}$

Answer 1

Answer: (D)

$\frac{3}{4}$

$Im\left(\frac{iz-2}{z-i}\right)+1=0$

Writing $z=x+iy$, we have

$Im\left(\frac{ix-y-2}{x+i(y-1)}\right)+1=0$

$\Rightarrow Im(\frac{ix-y-2}{x+i(y-1)}\times \frac{x-i(y-1)}{x-i(y-1)})+1=0$

$\Rightarrow Im\left(\frac{i{x}^2+i(y-1)(y+2)-x(y+2)+x(y-1)}{x^2+(y-1{)}^2}\right)+1=0$

$\Rightarrow \frac{x^2+y^2+y-2}{x^2+(y-1{)}^2}=-1$

$\Rightarrow x^2+y^2+y-2=-x^2-y^2+2y-1$

$\Rightarrow 2x^2+2y^2-y-1=0$

$\Rightarrow x^2+y^2-\frac{1}{2}y-\frac{1}{2}=0$

Comparing the equation with the general equation of a circle, we get

$\Rightarrow g=0,f=\frac{1}{4},c=-\frac{1}{2}$

Radius $=\sqrt{g^2+f^2-c}=\sqrt{\frac{1}{16}+\frac{1}{2}}=\sqrt{\frac{9}{16}}=\frac{3}{4}$

Hence, option $D$ is the correct answer.