A circle cuts a chord of length 4a on the x-a | vedclass

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(D) Let the center of the circle be $(h, k)$ and its radius be $r$.Since the circle cuts a chord of length $4a$ on the $x$-axis,the distance from the

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a parabola

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(D) Let the center of the circle be $(h, k)$ and its radius be $r$.Since the circle cuts a chord of length $4a$ on the $x$-axis,the distance from the center $(h, k)$ to the $x$-axis is $|k|$. By the property of a chord,$r^2 = k^2 + (2a)^2 = k^2 + 4a^2$.Since the circle passes through the point $(0, 2b)$ on the $y$-axis,the distance from the center $(h, k)$ to $(0, 2b)$ is $r$. Thus,$r^2 = (h - 0)^2 + (k - 2b)^2 = h^2 + (k - 2b)^2$.Equating the two expressions for $r^2$:$k^2 + 4a^2 = h^2 + (k - 2b)^2$$k^2 + 4a^2 = h^2 + k^2 - 4bk + 4b^2$$h^2 = 4bk - 4b^2 + 4a^2$$h^2 = 4b(k - b + a^2/b)$Replacing $(h, k)$ with $(x, y)$,the locus is $x^2 = 4b(y - b + a^2/b)$,which represents a parabola.

$A$ circle cuts a chord of length $4a$ on the $x$-axis and passes through a point on the $y$-axis,distant $2b$ from the origin. Then the locus of the center of this circle is

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