Equation of Circle Whose Extremities of a Diameter Given

Q. Let $P(x_1,y_1)$ and $Q(x_2,y_2)$ are two points such that their abscissa $x_1$ and $x_2$ are the roots of the equation $x^2+2x-3=0$ while the ordinates $y_1$ and $y_2$ are the roots of the equation $y^2+4y-12=0.$ The centre of the circle with PQ as diameter is

1. (-1, -2)
2. (1, 2)
3. (-1, 2)
4. (1, -2)

Q. The equation of circle whose diameter is the line joining the points (­–4, 3) and (12, –1) is

1. $x_2+(y_2+8x+2y+51=0$
2. $x_2+(y_2+8x-2y-51=0$
3. $x_2+(y_2+8x+2y-51=0$
4. $x_2+(y_2-8x-2y-51=0.$

Q. Equations of circles which pass through the points $(1,-2)$ and $(3,-4)$ and touch the $x-$axis is

1. $x^2+y^2+10x+20y+25=0$
2. $x^2+y^2+6x+2y+9=0$
3. $x^2+y^2-6x+4y+9=0$
4. $none$