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

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(D) Let $A \equiv (x_1, y_1)$ and $B \equiv (x_2, y_2)$.From the given equations,by the properties of roots of a quadratic equation:$x_1+x_2 = -2a$ an

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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 $B \equiv (x_2, y_2)$.From the given equations,by the properties of roots of a quadratic equation:$x_1+x_2 = -2a$ and $x_1x_2 = -b^2$.$y_1+y_2 = -2p$ and $y_1y_2 = -q^2$.The equation of a circle with diameter endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is given by $(x-x_1)(x-x_2) + (y-y_1)(y-y_2) = 0$.Expanding this,we get $x^2 - (x_1+x_2)x + x_1x_2 + y^2 - (y_1+y_2)y + y_1y_2 = 0$.Substituting the values of the sum and product of roots:$x^2 - (-2a)x + (-b^2) + y^2 - (-2p)y + (-q^2) = 0$.$x^2 + y^2 + 2ax + 2py - (b^2+q^2) = 0$.

The abscissae of the 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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