Explanation

(a)
Let $x_{1}$ and $x$ are the roots of the equation
$x^{2} + 2ax - b^{2} = 0$
$\therefore x_{1} + x_{2} = - 2a$ and $x_{1x_{2}} = - b^{2}$
Also, $y_{1}$ and $y_{2}$ are roots of the equation
$y^{2} + 2py - q^{2} = 0$
$\therefore\ \ y_{1} + y_{2} = - 2p$ and $y_{1}y_{2} = - q^{2}$
The equation of the circle with $P(x_{1},y_{1})$ and $Q(x_{2},y_{2})$ as then end points of diameter is
$\left( x - x_{1} \right)\left( x - x_{2} \right) + \left( y - y_{1} \right)\left( y - y_{2} \right) = 0$
$\Rightarrow x^{2} + y^{2} - x\left( x_{1} + x_{2} \right) - y\left( y_{1} + y_{2} \right) + x_{1}x_{2} + y_{1}y_{2} = 0$
$\Rightarrow x^{2} + y^{2} + 2ax + 2py - b^{2} - q^{2} = 0$