Let the abscissae of the two points P and Q on a circle be the...

Given that the roots of ${x}^{2} - 4x - 6 = 0$ are the abscissa of the end of diameter

i.e. ${x}_{1} + {x}_{2} = 4, {x}_{1}{x}_{2} = - 6$

and roots of ${y}^{2} + 2y - 7 = 0$ are ordinate of the end of diameter

i.e. ${y}_{1} + {y}_{2} = - 2, {y}_{1}{y}_{2} = - 7$

Now, equation of the circle will be

\left(x - {x}_{1}\right)\left(x - {x}_{2}\right) + \left(y - {y}_{1}\right)\left(y - {y}_{2}\right) = 0$</p><p>i.e.${x}^{2} - \left({x}_{1} + {x}_{2}\right)x + {x}_{1}{x}_{2} + {y}^{2} - \left({y}_{1} + {y}_{2}\right)y + {y}_{1}{y}_{2} = 0$</p><p>$⇒ {x}^{2} + {y}^{2} - 4x + 2y - 13 = 0$</p><p>$∴ a = - 2, b = 1, c = - 13$</p><p>$⇒ a + b - c = - 2 + 1 + 13 = 12