The abscissae of the two points A and B are the roots of the equation \(x^2+2 \mathrm{a} x-\mathrm{b}^2=0\) and their ordinates are roots of the equation \(y^2+2 p y-q^2=0\). Then the equation of the circle with 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,
\(\begin{aligned}
& x_1+x_2=-2 \mathrm{a}, x_1 x_2=-\mathrm{b}^2 \\
& y_1+y_2=-2 \mathrm{p}, y_1 y_2=-\mathrm{q}^2
\end{aligned}\)
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
\(\left(x-x_1\right)\left(x-x_2\right)+\left(y-y_1\right)\left(y-y_2\right)=0\)
\(\Rightarrow x^2+y^2-x\left(x_1+x_2\right)-y\left(y_1+y_2\right)+x_1 x_2 ~+\) \(y_1 y_2=0\)
\(\Rightarrow x^2+y^2+2 \mathrm{a} x+2 \mathrm{p} y-\mathrm{b}^2-\mathrm{q}^2=0\)