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(D) Let the equation of the circle be $x^2 + y^2 + 2gx + 2fy = 0$ as it passes through the origin $(0, 0)$.The circle cuts the $x$-axis at $A(-2g, 0)$

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$\frac{x_1}{x} + \frac{y_1}{y} = 2$

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(D) Let the equation of the circle be $x^2 + y^2 + 2gx + 2fy = 0$ as it passes through the origin $(0, 0)$.The circle cuts the $x$-axis at $A(-2g, 0)$ and the $y$-axis at $B(0, -2f)$.The equation of the line $AB$ is $\frac{x}{-2g} + \frac{y}{-2f} = 1$,which simplifies to $\frac{x}{g} + \frac{y}{f} = -2$.Since this line passes through the fixed point $(x_1, y_1)$,we have $\frac{x_1}{g} + \frac{y_1}{f} = -2$.The centre of the circle is $(-g, -f)$. Let the centre be $(x, y)$,so $x = -g$ and $y = -f$,which implies $g = -x$ and $f = -y$.Substituting these into the equation,we get $\frac{x_1}{-x} + \frac{y_1}{-y} = -2$.Multiplying by $-1$,we get $\frac{x_1}{x} + \frac{y_1}{y} = 2$.

$A$ circle passing through the origin cuts the coordinate axes at $A$ and $B$. If the straight line $AB$ passes through a fixed point $(x_1, y_1)$,then the locus of the centre of the circle is:

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