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Circle on PQ as diameter, where the coordinates of P and Q are roots of two quadratics

This type of question, asked 2 times: 25 Jun 2022, 26 Jul 2022. All 2 questions below, oldest first.

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  1. Question 1

    25 Jun 2022

    84
    Let the abscissae of the two points PP and QQ be the roots of 2x2−rx+p=02x^{2}- rx + p = 0 and the ordinates of PP and QQ be the roots of x2−sx−q=0x^{2}- sx - q = 0. If the equation of the circle described on PQPQ as diameter is 2(x2+y2)−11x−14y−22=02\left( x^{2}+y^{2} \right)- 11x - 14y - 22 = 0, then 2r+s−2q+p2r + s - 2q + p is equal to

  2. Question 2

    26 Jul 2022

    163
    Let the abscissae of the two points PP and QQ on a circle be the roots of x2−4x−6=0x^{2}- 4x - 6 = 0 and the ordinates of PP and QQ be the roots of y2+2y−7=y^{2}+ 2y - 7 = 0. If PQPQ is a diameter of the circle x2+y2+2ax+x^{2}+y^{2}+ 2ax + 2by+c=02by + c = 0, then the value of (a+b−c)(a + b - c) is

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