Find the equation of the circle with centre $(-2, 3)$ and radius $4$.

Let the abscissae of the two points $P$ and $Q$ be the roots of $2x^{2}-rx+p=0$ and the ordinates of $P$ and $Q$ be the roots of $y^{2}-sy-q=0$. If the equation of the circle described on $PQ$ as diameter is $2(x^{2}+y^{2})-11x-14y-22=0$,then $2r+s-2q+p$ is equal to

The sides of a rectangle are given by $x = \pm a$ and $y = \pm b$. Then the equation of the circle passing through the vertices of the rectangle is

The equation of the circle of radius $3$ that lies in the fourth quadrant and touches the lines $x=0$ and $y=0$ is

Circles are drawn through the point $(2,0)$ to cut intercepts of length $5$ units on the $X$-axis. If their centre lies in the first quadrant,then their equation is

$A$ circle passes through the points $(1,2)$ and $(3,4)$. If its centre lies on the line $x-y+3=0$,then its radius is equal to