If $(4,7)$ and $(-2,-1)$ are ends of a diameter of a circle which intersects $X$-axis at $A$ and $B$,then $AB$ is equal to

The abscissae of two points $P$ and $Q$ are the roots of the equation $2x^2 + 4x - 7 = 0$ and their ordinates are the roots of the equation $3x^2 - 12x - 1 = 0$. Then the centre of the circle with $PQ$ as a diameter is

The equation of the circle which passes through the points $(3, -2)$ and $(-2, 0)$ and whose centre lies on the line $2x - y = 3$ is:

Let in an equilateral $\Delta ABC,$ $A(-1 + a \cos \theta, 2 + a \sin \theta),$ $B(-1 + a \cos \alpha, 2 - a \sin \alpha),$ and $C(-1 + a \sin \beta, 2 + a \cos \beta).$ If the length of the median through vertex $A$ is $2b,$ then the equation of the circumcircle of triangle $ABC$ is (where $a$ is a constant) -

Find the equation of the circle with center $(2, 1)$ and touching the $X$-axis.

If the lines $3x - 4y - 7 = 0$ and $2x - 3y - 5 = 0$ are diameters of a circle with an area of $49\pi$ square units,then the equation of the circle is: