The locus of the centre of the circle which c | vedclass

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(C) Let the equation of the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$. The centre of the circle is $(-g, -f)$.The length of the intercept cut off by t

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

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(C) Let the equation of the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$. The centre of the circle is $(-g, -f)$.The length of the intercept cut off by the circle on the $x$-axis is $2\sqrt{g^2 - c} = 2a$,which implies $g^2 - c = a^2$  $(i)$.The length of the intercept cut off by the circle on the $y$-axis is $2\sqrt{f^2 - c} = 2b$,which implies $f^2 - c = b^2$  $(ii)$.Subtracting $(ii)$ from $(i)$,we get $g^2 - f^2 = a^2 - b^2$.Replacing $(-g, -f)$ with $(x, y)$,we have $g = -x$ and $f = -y$.Substituting these into the equation,we get $(-x)^2 - (-y)^2 = a^2 - b^2$,which simplifies to $x^2 - y^2 = a^2 - b^2$.

The locus of the centre of the circle which cuts off intercepts of length $2a$ and $2b$ from the $x$-axis and $y$-axis respectively,is

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