Question

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

• A. $\sqrt{(a+\overline{a})-r}$
• B. $\sqrt{(a+\overline{a}{)}^2-2r}$
• C. $\sqrt{(a+\overline{a}{)}^2-4r}$
• D. $\sqrt{(a+\overline{a}{)}^2-8r}$

Answer 1

The correct option is C $\sqrt{(a+\overline{a}{)}^2-4r}$

Points where the circle cuts the $x-$axis $z=\overline{z}.$
Hence subsituting $z=\overline{z}$ in the equation of circle, we get
$z^2+\overline{a}z+az+r=0$
$\Rightarrow z^2+(a+\overline{a})z+r=0$
$\Rightarrow AB=|z_1-z_2$|
(where $A$ and $B$ are points of intersectoin of circle with $x-$ axis)
$=\sqrt{(z_1+z_2{)}^2-4z_1{z}_2}$
$=\sqrt{(a+\overline{a}{)}^2-4r}$