Question

If $\lambda$ be real, prove that the equation ${|z-a|}^2+{|z-b|}^2=\lambda$ represents a circle. Determine its centre and radius.

Answer 1

${|z-a|}^2=(z-a)(\overline{z}-\overline{a})$
$=z\overline{z}+{\left|a\right|}^2-(z\overline{a}+\overline{z}a)$
$={\left|z\right|}^2+{\left|a\right|}^2-(z\overline{a}+\overline{z\overline{a}})$
$={\left|z\right|}^2+{\left|a\right|}^2-2Re\left(z\overline{a}\right)$ ......$\left(1\right)$
$\therefore L.H.S.$ $=2{\left|z\right|}^2-2Re(z\overline{a}+z\overline{b})+{\left|a\right|}^2+{\left|b\right|}^2=\lambda$
or ${\left|z\right|}^2-2Re\left[\frac{z(\overline{a}+\overline{b})}{2}\right]=\frac{1}{2}\{\lambda -{\left|a\right|}^2-{\left|b\right|}^2\}$
Add $\frac{{|a+b|}^2}{4}$ on both sides, then by $\left(1\right)$
$L.H.S.$ $=|z-\frac{{a+b}^2}{2}|=\frac{1}{2}\{2\lambda -{\left|a-b\right|}^2\}$
Above represents circle whose center is $\frac{a+b}{2}$ and radius is $\frac{1}{2}{\{2\lambda -{|a-b|}^2\}}^{1/2}$.