A line segment of $2$ units is sliding with its ends on two perpendicular lines. Then the locus of the middle point, is

x + 2 y + 1 = 0

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x^{2} + y^{2} = 1

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y^{2} = 4 a x

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x^{2} = 4 a y

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Solution

The correct option is B $x^{2} + y^{2} = 1$
Let $(h, k)$ be the middle point of $(0, a)$ and $(b, 0)$
$∴ a = 2 k$ and $b = 2 h$
Using distance formula we have
$\sqrt{(2 h)^{2} + (2 k)^{2}} = 2$
Squaring on both the sides, we have
$(2 h)^{2} + (2 k)^{2} = 4$
$h^{2} + k^{2} = 1$
Locus of the point will be $x^{2} + y^{2} = 1$

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