Question

Let P and Q be two points denoting the complex numbers $\alpha and\beta$ respectively on the complex plane. Which of the following equations can represent the equation of the circle passing through P and Q with least possible area ?

• A. $arg\left(\frac{z-\alpha }{z-\beta }\right)=\frac{\pi }{2}$
• B. $Re(z-\alpha )\left(\overline{z-\beta }\right)=0$
• C. $|z-\alpha {|}^2+|z-\beta {|}^2=(\overline{\alpha -\beta }{)}^2$
• D. $z\overline{z}+\left(\frac{\overline{\alpha }+\overline{\beta }}{2}\right)z+\left(\frac{\alpha +\beta }{2}\right)\overline{z}+\alpha \overline{\beta }+\overline{\alpha }\beta =0$

Answer 1

Answer: (B), (C)

The correct options are
B $Re(z-\alpha )\left(\overline{z-\beta }\right)=0$
C $|z-\alpha {|}^2+|z-\beta {|}^2=(\overline{\alpha -\beta }{)}^2$

(a) option is not correct as it should be $\pm \frac{\pi }{2}$
(b) Equation of circle is $\frac{z-\alpha }{z-\beta }$ is purely imaginary
$\therefore Re\left(\frac{z-\alpha }{z-\beta }\right)=0or\therefore Re(z-\alpha )\left(\overline{z-\beta }\right)=0\Rightarrow \left(b\right)iscorrect$
(c) Option is obviously correct
(d) Option cannot be correct as the centre of required circle is $\left(\frac{\alpha +\beta }{2}\right)$ but from the given equation centre of the circle is coefficient $\overline{z}=-\left(\frac{\alpha +\beta }{2}\right)$