The abscissa of two points A and B are roots of $x^2+4x-45=0$  and ordinates of A, B are roots of   $y^2-7y+12=0$ . Then centre of circle for which A, B as diameter is

Let $A\left(x_1,y_1\right),B\left(x_2,y_2\right)$  be two points.

Since $x_1,x_2$  are the roots of $x^2+4x-45=0$

$\therefore$ Sum of roots, $x_1+x_2=-4$ ,   product of  roots $x_1{x}_2=-45$

Since ordinates $y_1,y_2$  be the roots of $y^2-7y+12=0$

$\therefore$ Sum of roots, $y_1+y_2=7$ , product of roots, $y_1{y}_2=12$

The  equation of  the circle with A,B as diameter is

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

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

$\Rightarrow x^2+y^2+4x-7y-33=0$

$\therefore$ centre= $\left(-2,\frac{7}{2}\right)$