The abscissae of two points A and B are the r | vedclass

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(D) Let $A \equiv (x_1, y_1)$ and $B \equiv (x_2, y_2)$.According to the given condition,the roots of $x^2+2ax-b^2=0$ are $x_1, x_2$,so $x_1+x_2 = -2a

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$x^2+y^2+2ax+2py-(b^2+q^2)=0$

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(D) Let $A \equiv (x_1, y_1)$ and $B \equiv (x_2, y_2)$.According to the given condition,the roots of $x^2+2ax-b^2=0$ are $x_1, x_2$,so $x_1+x_2 = -2a$ and $x_1x_2 = -b^2$.Similarly,the roots of $y^2+2py-q^2=0$ are $y_1, y_2$,so $y_1+y_2 = -2p$ and $y_1y_2 = -q^2$.The equation of the circle with $A(x_1, y_1)$ and $B(x_2, y_2)$ as the endpoints of the diameter is given by $(x-x_1)(x-x_2) + (y-y_1)(y-y_2) = 0$.Expanding this,we get $x^2 - x(x_1+x_2) + x_1x_2 + y^2 - y(y_1+y_2) + y_1y_2 = 0$.Substituting the values,we get $x^2 - x(-2a) - b^2 + y^2 - y(-2p) - q^2 = 0$.Therefore,the equation is $x^2+y^2+2ax+2py-(b^2+q^2) = 0$.

The abscissae of two points $A$ and $B$ are the roots of the equation $x^2+2ax-b^2=0$ and their ordinates are roots of the equation $y^2+2py-q^2=0$. Then the equation of the circle with $AB$ as diameter is given by

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