A rod of length ‘I’ sides with its ends on two perpendicular lines. Then the locus of its midpoint is

x^{2} + y^{2} = \frac{I^{2}}{4}

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x^{2} + y^{2} = \frac{I^{2}}{2}

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x^{2} - y^{2} = \frac{I^{2}}{4}

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x^{2} - y^{2} = \frac{I^{2}}{3}

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Solution

The correct option is A $x^{2} + y^{2} = \frac{I^{2}}{4}$

The correct answer is option (a). Check the video for the approach.

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