Question

The point in the set { $z$ is an element of $C:arg\left(\frac{\left(z-2\right)}{\left(z-6i\right)}\right)=\frac{\pi }{2}$}(where $C$ denotes the set of all complex number) lie on the curve which is :

• A. Circle
• B. Pair of lines
• C. Parabola
• D. Hyperbola

Answer 1

The correct option is A

Circle

Explanation for the correct option

Find the curve on which the given point lies:

Given:

$arg\left(\frac{\left(z-2\right)}{\left(z-6i\right)}\right)=\frac{\pi }{2}$

$\Rightarrow arg\left(z-2\right)-arg\left(z-6i\right)=\frac{\pi }{2}$

We know, $z=x+iy$

We can write

$arg\left(\left(x-2\right)+iy\right)-arg\left(x+\left(y-6\right)i\right)=\frac{\pi }{2}$

$\Rightarrow {\tan }^{-1}\left(\frac{y}{\left(x-2\right)}\right)-{\tan }^{-1}\left(\frac{\left(y-6\right)}{x}\right)=\frac{\pi }{2}$

$\Rightarrow {\tan }^{-1}\left(\frac{{\displaystyle \frac{y}{\left(x-2\right)}-\frac{\left(y-6\right)}{x}}}{1+{\displaystyle \frac{y\left(y-6\right)}{x\left(x-2\right)}}}\right)=\frac{\pi }{2}\left[\because ,{\tan }^{-1}\left(A\right)-{\tan }^{-1}\left(B\right)={\tan }^{-1}\left(\frac{A-B}{1+\mathrm{AB}}\right)\right]$

$\Rightarrow$ $\left(\frac{{\displaystyle \frac{xy-\left(x-2\right)\left(y-6\right)}{x\left(x-2\right)}}}{{\displaystyle \frac{x\left(x-2\right)+y\left(y-6\right)}{x\left(x-2\right)}}}\right)=\tan \left(\frac{\pi }{2}\right)$

$\Rightarrow$ $\left(\frac{{\displaystyle xy-\left(x-2\right)\left(y-6\right)}}{x\left(x-2\right)+y\left(y-6\right)}\right)=\tan \left(\frac{\pi }{2}\right)$

$\Rightarrow$ $\left(\frac{{\displaystyle xy-\left(x-2\right)\left(y-6\right)}}{x\left(x-2\right)+y\left(y-6\right)}\right)=\frac{1}{0}$

$\Rightarrow$ $x\left(x-2\right)+y\left(y-6\right)=0$

$\Rightarrow$ $x^2-2x+1+y^2-6y+9-10=0$

$\Rightarrow$ ${\left(x-1\right)}^2+{\left(y-3\right)}^2={\left(\sqrt{10}\right)}^2$

Therefore, The point lies on a circle.

Hence option $A$ is the answer.