Let the abscissae of the two points P and Q b | vedclass

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(D) Let the points be $P(x_{1}, y_{1})$ and $Q(x_{2}, y_{2})$.The abscissae $x_{1}, x_{2}$ are roots of $2x^{2}-rx+p=0$,so $x_{1}+x_{2} = \frac{r}{2}$

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$7$

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(D) Let the points be $P(x_{1}, y_{1})$ and $Q(x_{2}, y_{2})$.The abscissae $x_{1}, x_{2}$ are roots of $2x^{2}-rx+p=0$,so $x_{1}+x_{2} = \frac{r}{2}$ and $x_{1}x_{2} = \frac{p}{2}$.The ordinates $y_{1}, y_{2}$ are roots of $y^{2}-sy-q=0$,so $y_{1}+y_{2} = s$ and $y_{1}y_{2} = -q$.The equation of the circle with $PQ$ as diameter is $(x-x_{1})(x-x_{2}) + (y-y_{1})(y-y_{2}) = 0$.$x^{2} - (x_{1}+x_{2})x + x_{1}x_{2} + y^{2} - (y_{1}+y_{2})y + y_{1}y_{2} = 0$.Substituting the values: $x^{2} - \frac{r}{2}x + \frac{p}{2} + y^{2} - sy - q = 0$.Multiplying by $2$: $2(x^{2}+y^{2}) - rx - 2sy + p - 2q = 0$.Comparing this with the given equation $2(x^{2}+y^{2}) - 11x - 14y - 22 = 0$:$r = 11$,$2s = 14 \implies s = 7$,and $p-2q = -22$.We need to find $2r+s-2q+p = 2(11) + 7 + (-22) = 22 + 7 - 22 = 7$.

Let the abscissae of the two points $P$ and $Q$ be the roots of $2x^{2}-rx+p=0$ and the ordinates of $P$ and $Q$ be the roots of $y^{2}-sy-q=0$. If the equation of the circle described on $PQ$ as diameter is $2(x^{2}+y^{2})-11x-14y-22=0$,then $2r+s-2q+p$ is equal to

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