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Q.

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

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answer is 7.

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Detailed Solution

The equation on of the circle having $P Q$ as diameter is $2 x^{2} - r x + p + 2 y^{2} - 2 s y - z q = 0$

$i . e ., 2 (x^{2} + y^{2}) - r x - 2 s y + p - z q = 0$

This comparing with $2 (x^{2} + y^{2}) - 11 x - 14 y - 22 = 0$

$r = 11, s = 7, p - 2 q = - 22$

$∴ 2 r + s - 2 q + p = 22 + 7 - 22 = 7$

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Let the abscisse of the two points P and Q be the roots of 2x2−rx+p=0 and the ordinates of P and Q be the roots of x2−sx−q=0 It the equation of the circle described on PQ as diameter is 2(x2+y2)−11x−14y−22=0, then 2r+s−2q+p is equal to_________