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

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

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

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

answer is D.

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Detailed Solution

Let the coordinates of $A$ and $B$ be $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ respectively. Then, $(x_{1}, y_{1})$ are roots of $x^{2} + 2 a x - b^{2} = 0$ and $(y_{1}, y_{2})$ are roots of

$x^{2} + 2 p x - q^{2} = 0$ .

$x_{1} + x_{2} = - 2 a, x_{1} x_{2} = - b^{2}$

and $y_{1} + y_{2} = - 2 p, y_{1} y_{2} = - q^{2}$

Now,

$\begin{array}{l} Radius = \frac{1}{2} A B \\ \Rightarrow Radius = \frac{1}{2} \sqrt{{(x_{2} - \bar{x_{1}})}^{2} + {(y_{2} - y_{1})}^{2}} \\ \Rightarrow Radius = \frac{1}{2} \sqrt{{(x_{1} + x_{2})}^{2}} + \bar{{(y_{1} + y_{2})}^{2} - 4 x_{1} x_{2} - 4 y_{1} y_{2}} \end{array}$

$\begin{aligned} \Rightarrow Radius = \frac{1}{2} \sqrt{4 a^{2} + 4 p^{2} + 4 b^{2} + 4 q^{2}} \\ = \sqrt{a^{2} + b^{2} + p^{2} + q^{2}} \end{aligned}$