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,we have:$x_1 + x_2 = -2a$ and $x_1 x_2 = -b^2$$y_1 + y

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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,we have:$x_1 + x_2 = -2a$ and $x_1 x_2 = -b^2$$y_1 + y_2 = -2p$ and $y_1 y_2 = -q^2$The equation of a circle with diameter $AB$ is given by $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$.Expanding this,we get: $x^2 - (x_1 + x_2)x + x_1 x_2 + y^2 - (y_1 + y_2)y + y_1 y_2 = 0$.Substituting the values:$x^2 - (-2a)x + (-b^2) + y^2 - (-2p)y + (-q^2) = 0$$x^2 + y^2 + 2ax + 2py - b^2 - q^2 = 0$.

The abscissae of $A$ 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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