Question

If the circles $z\overline{z}+\overline{{\alpha }_1}z+{\alpha }_1\overline{z}+{\alpha }_1\overline{z}+b_1=0$ & $z\overline{z}+\overline{{\alpha }_2}z+{\alpha }_2\overline{z}+{\alpha }_2\overline{z}+b_2=0$ where $b_1,b_2ϵR$ intersect orthogonally, then?

• A. ${\alpha }_1{\alpha }_2+\overline{{\alpha }_1}\overline{{\alpha }_2}=b_1+b_2$
• B. ${\alpha }_1{\alpha }_2-\overline{{\alpha }_1}\overline{{\alpha }_2}=b_1+b_2$
• C. ${\alpha }_1\overline{{\alpha }_2}+\overline{{\alpha }_1}{\alpha }_2=b_1+b_2$
• D. ${\alpha }_1\overline{{\alpha }_2}-\overline{{\alpha }_1}{\alpha }_2=b_1-b_2$

Answer 1

The correct option is C ${\alpha }_1\overline{{\alpha }_2}+\overline{{\alpha }_1}{\alpha }_2=b_1+b_2$

Given circles are $z\overline{z}+{\alpha }_{1}z+{\alpha }_1\overline{z}+b_1=0$ & $z\overline{z}+{\alpha }_{2}z+{\alpha }_2\overline{z}+b_2=0\therefore$ Centre $C_1=(-{\alpha }_1),$
Radius $R_1=\sqrt{{\left|{\alpha }_1\right|}^2-b_1}$ Centre $C_2=(-{\alpha }_2),$ Radius $R_2=\sqrt{{\left|{\alpha }_2\right|}^2-b_2}$ Now using the condition fur which circles are orthogonal
${\left|C_1{C}_2\right|}^2=R_1^2+R_2^2$ $\Rightarrow {|{\alpha }_1-{\alpha }_2|}^2={\left|{\alpha }_1\right|}^2-b_1+{\left|{\alpha }_2\right|}^2-b_2$ $\Rightarrow {\left|{\alpha }_1\right|}^2+{\left|{\alpha }_2\right|}^2-({\alpha }_1\overline{{\alpha }_2}+{\alpha }_2\overline{{\alpha }_1})={\left|{\alpha }_1\right|}^2+{\left|{\alpha }_2\right|}^2-(b_1+b_2)$
$\Rightarrow {\alpha }_1\overline{{\alpha }_2}+{\alpha }_2\overline{{\alpha }_1}=b_1+b_2$