Question

Locus of point $z$ so that $z,i,$ and $iz$ are collinear, is

• A. A straight line
• B. A circle
• C. An ellipse
• D. A rectangular hyperbola

Answer 1

The correct option is B A circle

Let $z=x+iy=(x,y)$
$i=(0,1)$
$iz=(-y,x)$
Hence the area has to be 0.
$\Rightarrow \left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|$

$\therefore \left|\begin{array}{ccc}0& 1& 1\\ x& y& 1\\ -y& x& 1\end{array}\right|$
$=-\left(1\right)(x+y)+\left(1\right)(x^2+y^2)$
$=0$ ----(Since the points are collinear, area is $0$).
$\Rightarrow x^2+y^2-x-y=0$
$\Rightarrow (x-\frac{1}{2}{)}^2+(y-\frac{1}{2}{)}^2-\frac{1}{2}=0$
$\Rightarrow (x-\frac{1}{2}{)}^2+(y-\frac{1}{2}{)}^2=\frac{1}{2}$
Hence $z$ lies on a circle.