The equation of the circle passing through the origin,whose center lies in the first quadrant and which makes intercepts of length $6$ and $4$ on the $x$-axis and $y$-axis respectively,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 four distinct points $(0,0), (2,0), (0,-2)$ and $(k,-2)$ are concyclic,if $k$ is equal to

$A$ circle passing through $(0,0)$,$(2,6)$,and $(6,2)$ cuts the $x$-axis at the point $P \neq (0,0)$. Then,the length of $OP$,where $O$ is the origin,is

Find the equation of the circle passing through the points $(2, 3)$ and $(-1, 1)$ and whose centre is on the line $x - 3y - 11 = 0$.

$A$ circle passes through the points $(2,0)$ and $(1,2)$. If the power of the point $(0,2)$ with respect to this circle is $4$,then the radius of the circle is