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Circle question

2022 · 25 Jun · Shift 1 · Q41

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JEE Main
Mathematics
Circle
2022 · 25 Jun · Shift 1 · Q41

Circle question

2022 · 25 Jun · Shift 1 · Q41

JEE MainMathematicsCircleNumerical+4 / −1

Let the abscissae of the two points P and Q be the roots of

2{x^2} - rx + p = 0

and the ordinates of P and Q be the roots of

{x^2} - sx - 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

\underline{\hspace{2cm}}

.

Numerical answer

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