locus of point for which the sum of squares of distance from the coordinates axes is 8 units is

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Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{Let} \mathrm{required} \mathrm{point} \mathrm{P}\hspace{0.17em}\mathrm{be} \mathrm{P}\left(\mathrm{h},\mathrm{k}\right) \\ \mathrm{According} \mathrm{to} \mathrm{question} \\ \mathrm{Distance} \mathrm{from} \mathrm{x}-\mathrm{axis}=\mathrm{PX}=\left|\mathrm{k}\right| \\ \mathrm{Distance} \mathrm{from} \mathrm{y}-\mathrm{axis}=\mathrm{PY}=\left|\mathrm{h}\right| \\ \mathrm{According} \mathrm{to} \mathrm{question} \\ {\mathrm{PY}}^2+{\mathrm{PX}}^2=8 \\ \Rightarrow {\mathrm{h}}^2+{\mathrm{k}}^2=8 \\ \mathrm{Put} \mathrm{h}\to \mathrm{x} \mathrm{and} \mathrm{k}\to \mathrm{y} \\ \Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2=8 \end{gathered}$$

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