Mathematics - Circle Question with Solution | TestHub

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

$\left(x-x_1\right)\left(x-x_2\right)+\left(y-y_1\right)\left(y-y_2\right)=0$

i.e. $x^2-\left(x_1+x_2\right)x+x_1{x}_2+y^2-\left(y_1+y_2\right)y+y_1{y}_2=0$

$\Rightarrow x^2+y^2-4x+2y-13=0$

$\therefore a=-2, b=1, c=-13$

$\Rightarrow a+b-c=-2+1+13=12$