The locus of a point P(x, y) satisfying the e | vedclass

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(C) The given equation is $\sqrt{(x-2)^2+y^2} + \sqrt{(x+2)^2+y^2} = 4$.Let $A = (2, 0)$ and $B = (-2, 0)$.The equation represents the sum of the dist

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a line segment

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(C) The given equation is $\sqrt{(x-2)^2+y^2} + \sqrt{(x+2)^2+y^2} = 4$.Let $A = (2, 0)$ and $B = (-2, 0)$.The equation represents the sum of the distances of point $P(x, y)$ from points $A$ and $B$,which is $PA + PB = 4$.The distance between points $A(2, 0)$ and $B(-2, 0)$ is $AB = \sqrt{(2 - (-2))^2 + (0 - 0)^2} = \sqrt{4^2} = 4$.Since $PA + PB = AB$,the point $P$ must lie on the line segment joining $A$ and $B$.Therefore,the locus of point $P(x, y)$ is the line segment connecting $(-2, 0)$ and $(2, 0)$.Thus,option $C$ is correct.

The locus of a point $P(x, y)$ satisfying the equation $\sqrt{(x-2)^2+y^2}+\sqrt{(x+2)^2+y^2}=4$ is:

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