JEE Main 2022 · 25 Jun · Shift 1 · Mathematics Q24

← JEE Main 2022 · 25 Jun · Shift 1

MathematicsInteger+4 / -1Question 24 of 90

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 $x^2-sx-q=0$ . If the equation of the circle described on PQ as diameter is $2\left(x^2+y^2\right)-11x-14y-22=0$ , then $2r+s-2q+p$ is equal to ______.

Your answer

Answer and worked solution

Official answer: 7

Diameter form. $(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0$ , i.e. $x^2-\frac r2x+\frac p2+y^2-sy-q=0$ .

Compare with $x^2+y^2-\frac{11}2x-7y-11=0$ : $r=11$ , $s=7$ , $\frac p2-q=-11$ .

Value. $2r+s-2q+p=22+7+2\left(\frac p2-q\right)=29-22=7$ .

Answer: 7

Topics:Equation of a circle Roots and coefficients

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JEE Main 2022 · 25 Jun · Shift 1, Mathematics question 24