Question

The circles $z\overline{z}+z\overline{a_1}+a_1\overline{z}+b_1=0,b_1\in R$ and $z\overline{z}+z\overline{a_2}+\overline{z_2}{a}_2+b_2=0,b_2\in R$ will intersect orthogonally if

• A. $2\text{Im}\left(b_1\overline{b_2}\right)=a_1+a_2$
• B. $2\text{Im}\left(\overline{b_1}{b}_2\right)=a_1+a_2$
• C. $2\text{Re}\left(a_2\overline{a_1}\right)=b_1+b_2$
• D. $2\text{Re}\left(a_1\overline{a_2}\right)=b_1+b_2$

Answer 1

The correct option is D $2\text{Re}\left(a_1\overline{a_2}\right)=b_1+b_2$

Centre and radius of
$z\overline{z}+z\overline{a_1}+\overline{z}{a}_1+b_1=0$ are $-a_1\text{and}\sqrt{a_1\overline{a_1}-b_1}$, respectively,
and that for other circle are $-a_2$ and $\sqrt{a_2\overline{a_2}-b_2}$, respectively.

$\because$These circles intersect orthogonally,
$\Rightarrow (C_1{C}_2{)}^2=r_1^2+r_2^2$
$\therefore |a_1-a_2{|}^2=a_1\overline{a_1}-b_1+a_2\overline{a_2}-b_2$
$\Rightarrow |a_1{|}^2+|a_2{|}^2+2\text{Re}\left(a_1\overline{a_2}\right)=|a_1{|}^2+|a_2{|}^2+b_1+b_2$
$\Rightarrow 2\text{Re}\left(a_1\overline{a_2}\right)=b_1+b_2$