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

Let the equation $ax^2+2hxy+by^2+2gx+2fy+c=0$ represent a point circle other than the origin. Then which one of the following conditions must hold?

The equation obtained by transforming $x^2+y^2-6x+10y-2=0$ to the parallel axes through $(3,-5)$ 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) -

The equation of the circle having its centre on the line $2x + y + 3 = 0$ and having the lines $3x + 4y - 18 = 0$ and $3x + 4y + 2 = 0$ as tangents is:

$A$ circle passes through the point $(1, -2)$ and touches the $x$-axis at $(3, 0)$. Through which of the following points does it also pass?