If the abscissa and ordinates of two points $P$ and $Q$ are the roots of the equations $x^{2} + 2 a x - b^{2} = 0$ and $x^{2} + 2 p x - q^{2} = 0$ respectively, then the equation of the circle with $P Q$ as diameter is

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Solution

Let $P (x_{1}, y_{1}), Q (x_{2}, y_{2})$ be end points of diameter.
Given
$x_{1}, x_{2}$ are the roots of $x^{2} + 2 a x - b^{2} = 0$
$y_{1}, y_{2}$ are the roots of $x^{2} + 2 p x - q^{2} = 0$
$\Rightarrow x_{1} + x_{2} = - 2 a, x_{1} x_{2} = - b^{2}$
$\Rightarrow y_{1} + y_{2} = - 2 p, y_{1} y_{2} = - q^{2}$
Equation of circle having $P Q$ as diameter is
$(x - x_{1}) (x - x_{2}) + (y - y_{1}) (y - y_{2}) = 0$
$\Rightarrow x^{2} - (x_{1} + x_{2}) x + x_{1} x_{2} + y^{2} - (y_{1} + y_{2}) y + y_{1} y_{2} = 0$
$\Rightarrow x^{2} + y^{2} + 2 a x + 2 p y - b^{2} - q^{2} = 0$

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The abscissae of the two points A and B are the roots of the equation $x^{2} + 2 a x - b^{2} = 0$ and their ordinates are the roots of the equation $x^{2} + 2 p x - q^{2} = 0$ . Find the equation of the circle with AB as diameter. Also, find its radius.