A line segment AM = a moves in the XOY plane | vedclass

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(B) Let the coordinates of point $A$ be $(a \cos 	heta, a \sin 	heta)$ since it lies on the circle $x^2 + y^2 = a^2$.Since $AM$ is parallel to the $

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$x^2 + y^2 = 2ax$

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(B) Let the coordinates of point $A$ be $(a \cos 	heta, a \sin 	heta)$ since it lies on the circle $x^2 + y^2 = a^2$.Since $AM$ is parallel to the $X$-axis and has length $a$,the coordinates of $M$ can be represented as $(x, y) = (a \cos 	heta + a, a \sin 	heta)$ or $(a \cos 	heta - a, a \sin 	heta)$.Considering the case $(x, y) = (a \cos 	heta + a, a \sin 	heta)$,we have $x - a = a \cos 	heta$ and $y = a \sin 	heta$.Squaring and adding these equations:$(x - a)^2 + y^2 = a^2 \cos^2 	heta + a^2 \sin^2 	heta$$(x - a)^2 + y^2 = a^2$$x^2 - 2ax + a^2 + y^2 = a^2$$x^2 + y^2 = 2ax$.Similarly,for the other case,we get $x^2 + y^2 = -2ax$.

$A$ line segment $AM = a$ moves in the $XOY$ plane such that $AM$ is parallel to the $X$-axis. If $A$ moves along the circle $x^2 + y^2 = a^2$,then the locus of $M$ is

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