The abscissae 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 $A B$ as diameter is

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

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

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

none of these

answer is A.

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Detailed Solution

Let the coordinates of $A$ and $B$ be $(x_{1}, y_{1})$

and $(x_{2}, y_{2})$ respectively. From the given

conditions, we then have $x_{1} + x_{2} = - 2 a, x_{1} x_{2} = - b^{2}, y_{1} + y_{2} = - 2 p$

and $y_{1} y_{2} = - q^{2}$ . Therefore the required radius is

$\begin{aligned} A B = (\frac{1}{2}) \sqrt{{(x_{1} - x_{2})}^{2} + {(y_{1} - y_{2})}^{2}} \\ r = (\frac{1}{2}) \sqrt{{(x_{1} + x_{2})}^{2} + {(y_{1} + y_{2})}^{2} - 4 (x_{1} x_{2} + y_{1} y_{2})} \\ r = (\frac{1}{2}) \sqrt{4 a^{2} + 4 p^{2} + 4 (b^{2} + q^{2})} \\ r = \sqrt{a^{2} + b^{2} + p^{2} + q^{2}} . \end{aligned}$