The centre of the circle $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$ is

The lines $2x - 3y = 5$ and $3x - 4y = 7$ are diameters of a circle with an area of $154$ square units. Find the equation of the circle.

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) -

The centre of the circle $r^2-4r(\cos \theta+\sin \theta)-4=0$ in Cartesian coordinates is

The equation of the circle concentric with the circle $x^2 + y^2 - 6x + 7 = 0$ and which touches the line $x + y + 3 = 0$ is:

If the equation of the circle passing through the point $(8,8)$ and having the lines $x+2y-2=0$ and $2x+3y-1=0$ as its diameters is $x^2+y^2+px+qy+r=0$,then $p^2+q^2+r=$