If sum of square of distance of a point from axes is $4$ , then its locus is

x + y = 2

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x^{2} + y^{2} = 16

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x + y = 4

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

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Solution

The correct option is D $x^{2} + y^{2} = 4$
$l e t t h e p o i n t b e (x, y), t h e n y^{2} + x^{2} = 4 a s d i s t a n c e o f p o i n t (x, y) f r o m X - a x i s a n d f r o m Y - a x i s = x u n i t$

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Similar questions

A point moves in such a way that the sum of square of its distance from the points A(2, 0) and B(-2, 0) is always

equal to the square of the distance between A and B. The locus of the point is

\frac{x^{2}}{y^{2}} + \frac{y^{2}}{x^{2}} - \frac{1}{i} (\frac{x}{y} - \frac{y}{x}) - \frac{9}{4}

x

y

x^{2} + y^{2} - 4 x - 4 y + 4 = 0

x + y - x y + k \sqrt{x^{2} + y^{2}} = 0

k

x = 2

x = 2

4

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