A circle is drawn to cut a chord of length 2a | vedclass

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(B) Let the centre of the circle be $O(x, y)$.Since the circle touches the $Y$-axis,the radius of the circle is $r = |x|$.The perpendicular drawn from

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$x^2 - y^2 = a^2$

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(B) Let the centre of the circle be $O(x, y)$.Since the circle touches the $Y$-axis,the radius of the circle is $r = |x|$.The perpendicular drawn from the centre $O(x, y)$ to the chord on the $X$-axis bisects the chord.Let the chord be $AB$ of length $2a$. The midpoint of the chord is $M(x, 0)$.The distance from the centre $O(x, y)$ to the midpoint $M(x, 0)$ is $|y|$.In the right-angled triangle formed by the centre,the midpoint of the chord,and one endpoint of the chord,the hypotenuse is the radius $r = |x|$.By Pythagoras theorem,$r^2 = |y|^2 + a^2$.Substituting $r = x$,we get $x^2 = y^2 + a^2$,which simplifies to $x^2 - y^2 = a^2$.

$A$ circle is drawn to cut a chord of length $2a$ units along the $X$-axis and to touch the $Y$-axis. The locus of the centre of the circle is

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