Let, $P({x_1},{y_1})$ and $Q({x_2},{y_2})$ are two points such that their abscissa ${x_1},{x_2}$ are the roots of the equation ${x^2} + 2x - 3 = 0$ while the ordinate ${y_1},{y_2}$ are the roots of the equation ${y^2} + 4y - 12 = 0$ . Then find the center of the circle with PQ as diameter.
A. (-1, -2)
B. (1,2)
C. (1, -2)
D. (-1,2)

Hint: First solve the given quadratic equation ${x^2} + 2x - 3 = 0$ to obtain the abscissa and then solve ${y^2} + 4y - 12 = 0$ to obtain the ordinates. Add -3 and 1 then divide -2 by 2 to obtain the abscissa of the center and add -6 and 2, then divide -4 by 2 to obtain the ordinate of the center of the circle.

Formula Used:
The middle point of the line AB, where $A(a,b)$ and $B(c,d)$, is $\left( {\dfrac{{a + c}}{2},\dfrac{{b + d}}{2}} \right)$.

Complete step by step solution:
Solve ${x^2} + 2x - 3 = 0$ to obtain the abscissa.
${x^2} + 3x - x - 3 = 0$
$x(x + 3) - 1(x + 3) = 0$
$(x - 1)(x + 3) = 0$
$x = 1, - 3$
Solve ${y^2} + 4y - 12 = 0$ to obtain the ordinates.
${y^2} + 6y - 2y - 12 = 0$
$y(y + 6) - 2(y + 6) = 0$
$(y - 2)(y + 6) = 0$
$y = 2, - 6$
Hence, the center of the circle is $\left( {\dfrac{{1 - 3}}{2},\dfrac{{2 - 6}}{2}} \right)$ .
That is $( - 1, - 2)$.

Option ‘A’ is correct

Note: Sometimes students get confused and obtain the line equation of PQ by the coordinates and then try to obtain the center, but that is not needed. Centre of a circle is always the middle point of any diameter, so apply the midpoint formula to obtain the required answer.