The locus of point of intersection of perpendicular straight lines which are at constant distance ‘p’ from origin is

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$x^{2} + y^{2} = 4 p^{2}$

$x^{2} + y^{2} = 3 p^{2}$

$x^{2} + y^{2} = 2 p^{2}$

$x^{2} + y^{2} = p^{2}$

answer is C.

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Detailed Solution

$\begin{array}{l} L e t P (x, y) b e t h e p o int o n l o c u s \\ I n Δ^{l e} O A P \\ O A^{2} + P A^{2} = O P^{2} \\ \Rightarrow p^{2} + p^{2} = x^{2} + y^{2} \\ \Rightarrow x^{2} + y^{2} = 2 p^{2} \end{array}$

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