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(D) Let the equation of the circle be $(x-h)^2 + (y-k)^2 = r^2$,where $(h, k)$ is the centre and $r$ is the radius.The intercept on the $X$-axis is gi

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$x^2-y^2-7=0$

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(D) Let the equation of the circle be $(x-h)^2 + (y-k)^2 = r^2$,where $(h, k)$ is the centre and $r$ is the radius.The intercept on the $X$-axis is given by $2\sqrt{r^2 - k^2} = 8$,which implies $r^2 - k^2 = 16$,so $r^2 = k^2 + 16$.The intercept on the $Y$-axis is given by $2\sqrt{r^2 - h^2} = 6$,which implies $r^2 - h^2 = 9$,so $r^2 = h^2 + 9$.Equating the two expressions for $r^2$,we get $k^2 + 16 = h^2 + 9$.Rearranging the terms,we get $h^2 - k^2 = 7$.Replacing $(h, k)$ with $(x, y)$,the locus of the centre is $x^2 - y^2 = 7$,or $x^2 - y^2 - 7 = 0$.

If the intercepts made by a variable circle on the $X$-axis and $Y$-axis are $8$ and $6$ units respectively,then the locus of the centre of the circle is

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