Question

The circles $z\overline{z}+z{\overline{a}}_1+z{\overline{a}}_1+b_1=0,b_1\in R$ and $z\overline{z}+z{\overline{a}}_2+z{\overline{a}}_2+b_2=0,b_2\in R$ intersects orthogonally, then prove that $Re\left(a_1{\overline{a}}_2\right)=b_1+b_2$.

Answer 1

Center and radius of $z\overline{z}+z{\overline{a}}_1+z{\overline{a}}_1+b_1=0$ are $-a_1$ 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. These circle will intersect orthogonally if the sum of squares of radii is equal to the square of distance between their centers. Therfore,
${|a_1-a_2|}^2=a_1{\overline{a}}_1-b_1+a_2{\overline{a}}_2-b_2$
$⟹a_1{\overline{a}}_1+a_2{\overline{a}}_2-a_1{\overline{a}}_2-{\overline{a}}_1{a}_2=a_1{\overline{a}}_1+a_1{\overline{a}}_2-b_1-b_2$
$⟹a_1{\overline{a}}_2+{\overline{a}}_1{a}_2=b_1+b_2$
$⟹2Re\left(a_1{\overline{a}}_2\right)=b_1+b_2$
Ans: 1