If the abscissae of points $A, B$ are the roots of the equation $x^{2} + 2 a x - b^{2} = 0$ and ordinates of $A, B$ are roots of $y^{2} + 2 p y - q^{2} = 0$ , then find the equation of a circle for which $¯ ¯¯¯¯¯¯ ¯ A B$ is a diameter

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Solution

Let $x_{1}, x_{2}$ be the roots of the equation $x^{2} + 2 a x - b^{2} = 0$ . Then
$x_{1} + x_{2} = - 2 a$ and $(x_{1} - x_{2}) = \sqrt{(x_{1} + x_{2})^{2} - 4 x_{1} x_{2}} = \sqrt{4 a^{2} + 4 b^{2}}$
Similarly
Let $y_{1}, y_{2}$ be the roots of the equation $x^{2} + 2 p x - q^{2} = 0$ . Then
$y_{1} + y_{2} = - 2 p$ and $(y_{1} - y_{2}) = \sqrt{(y_{1} + y_{2})^{2} - 4 y_{1} y_{2}} = \sqrt{4 p^{2} + 4 q^{2}}$
Hence the centre of the circle with diameter AB is
$C = (\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})$
$= (\frac{- 2 a}{2}, \frac{- 2 p}{2})$
$= (- a, - p)$
And the radius of the circle is given by
$\frac{A B}{2} = 0.5 \sqrt{(x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2}}$
$= 0.5 \sqrt{4 a^{2} - 4 b^{2} + 4 p^{2} - 4 q^{2}}$
$= \sqrt{a^{2} + p^{2} - b^{2} - q^{2}}$
Hence the equation of the circle is
$(x + a)^{2} + (y + p)^{2} = a^{2} + p^{2} - b^{2} - q^{2}$
$x^{2} + y^{2} + 2 a x + 2 p y + a^{2} + p^{2} = a^{2} + p^{2} - b^{2} - q^{2}$
$x^{2} + y^{2} + 2 a x + 2 p y + b^{2} + q^{2} = 0$

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If the abscissae of points A,B are the roots of the equation x2+2ax−b2=0 and ordinates of A,B are roots of y2+2py−q2=0, then find the equation of a circle for which ¯¯¯¯¯¯¯¯AB is a diameter

If the abscissae of points $A, B$ are the roots of the equation $x^{2} + 2 a x - b^{2} = 0$ and ordinates of $A, B$ are roots of $y^{2} + 2 p y - q^{2} = 0$ , then find the equation of a circle for which $¯ ¯¯¯¯¯¯ ¯ A B$ is a diameter