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QJEEM22Ma7957: JEE Main 2022 Mathematics - Class 11 JEEM22Ma7957 Solution: Let the abscissae of the two points P and Q be the roots of and …

Let the abscissae of the two points P and Q be the roots of 2x2−rx+p=02{x^2} - rx + p = 0 and the ordinates of P and Q be the roots of x2−sx−q=0{x^2} - sx - q = 0 . If the equation of the circle described on PQ as diameter is 2(x2+y2)−11x−14y−22=02({x^2} + {y^2}) - 11x - 14y - 22 = 0 , then 2r+s−2q+p2r + s - 2q + p is equal to __________.

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Let P(x1,y1)P({x_1},{y_1}) & Q(x2,y2)Q({x_2},{y_2})

∴\therefore Roots of 2x2−rx+p=02{x^2} - rx + p = 0 are x1, x2{x_1},\,{x_2}

and roots of x2−sx−q=0{x^2} - sx - q = 0 are y1, y2{y_1},\,{y_2} .

∴\therefore Equation of circle ≡(x−x1)(x−x2)+(y−y1)(y−y2)=0\equiv (x - {x_1})(x - {x_2}) + (y - {y_1})(y - {y_2}) = 0

⇒x2−(x1+x2)x+x1x2+y2−(y1+y2)y+y1y2=0\Rightarrow {x^2} - ({x_1} + {x_2})x + {x_1}{x_2} + {y^2} - ({y_1} + {y_2})y + {y_1}{y_2} = 0

⇒x2−r2x+p2+y2+sy−q=0\Rightarrow {x^2} - {r \over 2}x + {p \over 2} + {y^2} + sy - q = 0

⇒2x2+2y2−rx+2sy+p−2q=0\Rightarrow 2{x^2} + 2{y^2} - rx + 2sy + p - 2q = 0

Compare with 2x2+2y2−11x−14y−22=02{x^2} + 2{y^2} - 11x - 14y - 22 = 0

We get r=11, s=7, p−2q=−22r = 11,\,s = 7,\,p - 2q = - 22

⇒2r+s+p−2q=22+7−22=7\Rightarrow 2r + s + p - 2q = 22 + 7 - 22 = 7

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