prove that |z-z1|²+|z-z2|²=k will represent a circle  if |z1-z2|²<=2k

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$$\begin{gathered} {\left|\mathrm{z}-{\mathrm{z}}_1\right|}^2+{\left|\mathrm{z}-{\mathrm{z}}_2\right|}^2=\mathrm{k} \\ \mathrm{Put} \mathrm{z}=\mathrm{x}+\mathrm{iy} \\ {\mathrm{z}}_1={\mathrm{x}}_1+{\mathrm{iy}}_1 \\ {\mathrm{z}}_2={\mathrm{x}}_2+{\mathrm{iy}}_2 \\ \Rightarrow {\left|\mathrm{x}+\mathrm{iy}-{\mathrm{x}}_1-{\mathrm{iy}}_1\right|}^2+{\left|\mathrm{x}+\mathrm{iy}-{\mathrm{x}}_2-{\mathrm{iy}}_2\right|}^2=\mathrm{k} \\ \Rightarrow {\left|\left(\mathrm{x}-{\mathrm{x}}_1\right)+\mathrm{i}\left(\mathrm{y}-{\mathrm{y}}_1\right)\right|}^2+{\left|\left(\mathrm{x}-{\mathrm{x}}_2\right)+\mathrm{i}\left(\mathrm{y}-{\mathrm{y}}_2\right)\right|}^2=\mathrm{k} \\ \Rightarrow {\left\{\sqrt{{\left(\mathrm{x}-{\mathrm{x}}_1\right)}^2+{\left(\mathrm{y}-{\mathrm{y}}_1\right)}^2}\right\}}^2+{\left\{\sqrt{{\left(\mathrm{x}-{\mathrm{x}}_2\right)}^2+{\left(\mathrm{y}-{\mathrm{y}}_2\right)}^2}\right\}}^2=\mathrm{k} \\ \Rightarrow {\mathrm{x}}^2+{{\mathrm{x}}_1}^2-2{\mathrm{xx}}_1+{\mathrm{y}}^2+{{\mathrm{y}}_1}^2-2{\mathrm{yy}}_1+{\mathrm{x}}^2+{{\mathrm{x}}_2}^2-2{\mathrm{xx}}_2+{\mathrm{y}}^2+{{\mathrm{y}}_2}^2-2{\mathrm{yy}}_2=\mathrm{k} \\ \Rightarrow 2{\mathrm{x}}^2+2{\mathrm{y}}^2-2\mathrm{x}\left({\mathrm{x}}_1+{\mathrm{x}}_2\right)-2\mathrm{y}\left({\mathrm{y}}_1+{\mathrm{y}}_2\right)+{{\mathrm{x}}_1}^2+{{\mathrm{y}}_1}^2+{{\mathrm{x}}_2}^2+{{\mathrm{y}}_2}^2-\mathrm{k}=0 \\ \Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2-\mathrm{x}\left({\mathrm{x}}_1+{\mathrm{x}}_2\right)-\mathrm{y}\left({\mathrm{y}}_1+{\mathrm{y}}_2\right)+\frac{{{\mathrm{x}}_1}^2+{{\mathrm{y}}_1}^2+{{\mathrm{x}}_2}^2+{{\mathrm{y}}_2}^2-\mathrm{k}}{2}=0...\left(\mathrm{i}\right) \\ \mathrm{for} {\mathrm{x}}^2+{\mathrm{y}}^2+2\mathrm{gx}+2\mathrm{fy}+\mathrm{c}=0, \mathrm{we} \mathrm{have} \mathrm{centre} \mathrm{as} \left(-\mathrm{g},-\mathrm{f}\right) \mathrm{and} \mathrm{radius} \mathrm{as} \\ \sqrt{{\mathrm{g}}^2+{\mathrm{f}}^2-\mathrm{c}} \\ \mathrm{For} \left(\mathrm{i}\right) \\ \mathrm{g}=-\frac{\left({\mathrm{x}}_1+{\mathrm{x}}_2\right)}{2}, \mathrm{f}=-\frac{\left({\mathrm{y}}_1+{\mathrm{y}}_2\right)}{2} \mathrm{and} \mathrm{c}=\frac{{{\mathrm{x}}_1}^2+{{\mathrm{y}}_1}^2+{{\mathrm{x}}_2}^2+{{\mathrm{y}}_2}^2-\mathrm{k}}{2} \\ \mathrm{Radius} \mathrm{of} \mathrm{circle}\ge 0 \left[\mathrm{In} \mathrm{case} \mathrm{of} \mathrm{equlity}, \mathrm{we} \mathrm{get} \mathrm{point} \mathrm{circle}\right] \\ \sqrt{{\mathrm{g}}^2+{\mathrm{f}}^2-\mathrm{c}}\ge 0 \\ \Rightarrow {\mathrm{g}}^2+{\mathrm{f}}^2-\mathrm{c}\ge 0 \\ \Rightarrow {\left\{-\frac{\left({\mathrm{x}}_1+{\mathrm{x}}_2\right)}{2}\right\}}^2+{\left\{-\frac{\left({\mathrm{y}}_1+{\mathrm{y}}_2\right)}{2}\right\}}^2-\left\{\frac{{{\mathrm{x}}_1}^2+{{\mathrm{y}}_1}^2+{{\mathrm{x}}_2}^2+{{\mathrm{y}}_2}^2-\mathrm{k}}{2}\right\}\ge 0 \\ \Rightarrow \frac{{{\mathrm{x}}_1}^2+{{\mathrm{x}}_2}^2+2{\mathrm{x}}_1{\mathrm{x}}_2}{4}+\frac{{{\mathrm{y}}_1}^2+{{\mathrm{y}}_2}^2+2{\mathrm{y}}_1{\mathrm{y}}_2}{4}+\frac{-{{\mathrm{x}}_1}^2-{{\mathrm{y}}_1}^2-{{\mathrm{x}}_2}^2-{{\mathrm{y}}_2}^2+\mathrm{k}}{2}\ge 0 \\ \Rightarrow {{\mathrm{x}}_1}^2+{{\mathrm{x}}_2}^2+2{\mathrm{x}}_1{\mathrm{x}}_2+{{\mathrm{y}}_1}^2+{{\mathrm{y}}_2}^2+2{\mathrm{y}}_1{\mathrm{y}}_2-2{{\mathrm{x}}_1}^2-2{{\mathrm{y}}_1}^2-2{{\mathrm{x}}_2}^2-2{{\mathrm{y}}_2}^2+2\mathrm{k}\ge 0 \\ \Rightarrow 2{\mathrm{x}}_1{\mathrm{x}}_2+2{\mathrm{y}}_1{\mathrm{y}}_2- {{\mathrm{x}}_1}^2-{{\mathrm{y}}_1}^2-{{\mathrm{x}}_2}^2-{{\mathrm{y}}_2}^2+2\mathrm{k}\ge 0 \\ \Rightarrow {{\mathrm{x}}_1}^2+{{\mathrm{x}}_2}^2-2{\mathrm{x}}_1{\mathrm{x}}_2+{{\mathrm{y}}_1}^2+{{\mathrm{y}}_2}^2-2{\mathrm{y}}_1{\mathrm{y}}_2\le 2\mathrm{k} \\ \Rightarrow {\left({\mathrm{x}}_1-{\mathrm{x}}_2\right)}^2+{\left({\mathrm{y}}_1-{\mathrm{y}}_2\right)}^2\le 2\mathrm{k} \\ \Rightarrow {\left\{\sqrt{{\left({\mathrm{x}}_1-{\mathrm{x}}_2\right)}^2+{\left({\mathrm{y}}_1-{\mathrm{y}}_2\right)}^2}\right\}}^2\le 2\mathrm{k} \\ \mathrm{As} {\mathrm{z}}_1={\mathrm{x}}_1+{\mathrm{iy}}_1 \mathrm{and} {\mathrm{z}}_2={\mathrm{x}}_2+{\mathrm{iy}}_2 \mathrm{which} \mathrm{gives} {\mathrm{z}}_1-{\mathrm{z}}_2=\left({\mathrm{x}}_1-{\mathrm{x}}_2\right)+\mathrm{i}\left({\mathrm{y}}_1-{\mathrm{y}}_2\right), \mathrm{hence} \\ {\left\{\sqrt{{\left({\mathrm{x}}_1-{\mathrm{x}}_2\right)}^2+{\left({\mathrm{y}}_1-{\mathrm{y}}_2\right)}^2}\right\}}^2\le 2\mathrm{k} \mathrm{becomes} \\ {\left\{\left|{\mathrm{z}}_1-{\mathrm{z}}_2\right|\right\}}^2\le 2\mathrm{k} \\ \Rightarrow {\left|{\mathrm{z}}_1-{\mathrm{z}}_2\right|}^2\le 2\mathrm{k}.....\mathbf{Hence}\mathbf{ }\mathbf{proved} \end{gathered}$$

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