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(D) Let the equation of the circle be $x^2+y^2+2gx+2fy+c=0$. The center of the circle is $(-g, -f)$.The length of the intercept made by the circle on

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

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(D) Let the equation of the circle be $x^2+y^2+2gx+2fy+c=0$. The center of the circle is $(-g, -f)$.The length of the intercept made by the circle on the $x$-axis is $2\sqrt{g^2-c} = 2a$,which implies $g^2-c = a^2$,so $c = g^2-a^2$.The length of the intercept made by the circle on the $y$-axis is $2\sqrt{f^2-c} = 2b$,which implies $f^2-c = b^2$,so $c = f^2-b^2$.Equating the two expressions for $c$,we get $g^2-a^2 = f^2-b^2$,which simplifies to $g^2-f^2 = a^2-b^2$.Replacing $(-g, -f)$ with the coordinates $(x, y)$ of the center,we get the locus $x^2-y^2 = a^2-b^2$.

Two straight rods of lengths $2a$ and $2b$ move along the coordinate axes in such a way that their extremities are always concyclic. Then the locus of the centres of such circles is

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