The abscissae of A and B are the roots of the | vedclass

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(A) Let the coordinates of $A$ be $(x_1, y_1)$ and $B$ be $(x_2, y_2)$.From the given equations,the roots for abscissae are $x_1, x_2$,so $x_1 + x_2 =

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

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(A) Let the coordinates of $A$ be $(x_1, y_1)$ and $B$ be $(x_2, y_2)$.From the given equations,the roots for abscissae are $x_1, x_2$,so $x_1 + x_2 = -2a$ and $x_1x_2 = -b^2$.Similarly,the roots for ordinates are $y_1, y_2$,so $y_1 + y_2 = -2p$ and $y_1y_2 = -q^2$.The equation of a circle with diameter $AB$ 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,we get $x^2 - (-2a)x + (-b^2) + y^2 - (-2p)y + (-q^2) = 0$.This simplifies to $x^2 + y^2 + 2ax + 2py - b^2 - q^2 = 0$.

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

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