Inspired by Clement V Durell

Let the circle be modeled as $x^2 + y^2 = R^2$ and the point P be $P(-x_{0},0)$ . Let the chord QR be modeled by the linear equation $y - 0 = -(tan(\frac{\pi}{4}))(x + x_{0}) \Rightarrow y= -x-x_{0}.$ Substitution of this chord equation into the circular equation yields:

$x^2 + (-x-x_{0})^2 = R^2 \Rightarrow 2x^2 + 2x_{0}x + (x_{0}^2 - R^2) = 0$

and by the Quadratic Equation:

$x = \frac{-2x_{0} \pm \sqrt{4x_{0}^2 - 4(2)(x_{0}^2 - R^2)}}{4} = \frac{-x_{0} \pm \sqrt{2R^2 - x_{0}^2}}{2}$ ;

which gives the points:

$Q( \frac{-x_{0} - \sqrt{2R^2 - x_{0}^2}}{2}, \frac{-x_{0} + \sqrt{2R^2 - x_{0}^2}}{2})$ and $R( \frac{-x_{0} + \sqrt{2R^2 - x_{0}^2}}{2}, \frac{-x_{0} - \sqrt{2R^2 - x_{0}^2}}{2})$ .

Now, the squared distances $PQ^2, PR^2$ each compute to:

$PQ^2 = [\frac{-x_{0} - \sqrt{2R^2 - x_{0}^2}}{2} + x_{0}]^2 + [\frac{-x_{0} + \sqrt{2R^2 - x_{0}^2}}{2} - 0]^2 = 2 \cdot [\frac{x_{0} - \sqrt{2R^2 - x_{0}^2}}{2}]^2$ ;

$PR^2 = [\frac{-x_{0} + \sqrt{2R^2 - x_{0}^2}}{2} + x_{0}]^2 + [\frac{-x_{0} - \sqrt{2R^2 - x_{0}^2}}{2} - 0]^2 = 2 \cdot [\frac{x_{0} + \sqrt{2R^2 - x_{0}^2}}{2}]^2$ .

Finally, we calculate:

$\frac{AB^2}{PQ^2 + PR^2} = \frac{(2R)^2}{ 2 \cdot [\frac{x_{0} - \sqrt{2R^2 - x_{0}^2}}{2}]^2 +2 \cdot [\frac{x_{0} + \sqrt{2R^2 - x_{0}^2}}{2}]^2} = \frac{8R^2}{4R^2} = \boxed{2}.$ ;

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