Question

Intercept made by the circle $z\overline{z}+\overline{\alpha }z+\alpha \overline{z}+r=0$ on the real axis of complex plane is

• A. $\sqrt{(\alpha +\overline{\alpha })-r}$
• B. $\sqrt{(\alpha +\overline{\alpha }{)}^2-2r}$
• C. $\sqrt{(\alpha +\overline{\alpha }{)}^2-4r}$
• D. $\sqrt{(\alpha +\overline{\alpha }{)}^2-r}$

Answer 1

Answer: (C)

$\sqrt{(\alpha +\overline{\alpha }{)}^2-4r}$

Points where the circle cuts the axis $z=\overline{z}$.
Hence, substituting $z=\overline{z}$ in the equation of circle, we get
$z^2+\overline{\alpha }z+\alpha z+r=0$
or $z^2+(\alpha +\overline{\alpha })z+r=0$
$⟹AB=|z_1-z_2|$ (where A and B are points of intersection of circle with the x-axis)
$=\sqrt{(z_1+z_2{)}^2-4z_1{z}_2}$
$=\sqrt{(\alpha +\overline{\alpha }{)}^2-4r}$
Ans: C