The abscissae of two points $\mathrm{A}$ and $\mathrm{B}$ are the roots of the equation $x^2+2 a x-b^2=0$…
The abscissae of two points $\mathrm{A}$ and $\mathrm{B}$ are the roots of the equation $x^2+2 a x-b^2=0$ and their ordinates are roots of the equation $y^2+2 \mathrm{p} y-\mathrm{q}^2=0$. Then the equation of the circle with $\mathrm{AB}$ as diameter is given by
Let $\mathrm{A} \equiv\left(x_1, y_1\right)$ and $\mathrm{B} \equiv\left(x_2, y_2\right)$.
According to the given condition,
$x_1+x_2=-2 \mathrm{a}, x_1 x_2=-\mathrm{b}^2$
$y_1+y_2=-2 \mathrm{p}, y_1 y_2=-\mathrm{q}^2$
The equation of the circle with $\mathrm{A}\left(x_1, y_1\right)$ and $\mathrm{B}\left(x_2, y_2\right)$ as the end points of diameter is
$\left(x-x_1\right)\left(x-x_2\right)+\left(y-y_1\right)\left(y-y_2\right)=0$
$\Rightarrow x^2+y^2-x\left(x_1+x_2\right)-y\left(y_1+y_2\right)+x_1 x_2+y_1 y_2=0$
$\Rightarrow x^2+y^2+2 \mathrm{a} x+2 \mathrm{p} y-\mathrm{b}^2-\mathrm{q}^2=0$