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The abscissa of 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 $x^{2} + 2 p x - q^{2} = 0$ . The radius of the circle with AB as diameter is :

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$\sqrt{a^{2} + b^{2} + p^{2} + q^{2}}$

$\sqrt{a^{2} + p^{2}}$

$\sqrt{b^{2} + q^{2}}$

$\sqrt{a^{2} + b^{2} + p^{2} + 1}$

answer is A.

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Detailed Solution

$G i v e n e q u a t i o n s x^{2} + 2 a x - b^{2} = 0, x^{2} + 2 p x - q^{2} = 0$
$A d d t h e t w o e q u a t i o n s, w e w i l l g e t t h e r e q u i r e d c i r c l e e q u a t i o n x^{2} + 2 a x - b^{2} + y^{2} + 2 p y - q^{2} = 0$

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