If $x-y+1=0$ meets the circle $x^2+y^2+y-1=0$ at $A$ and $B$,then the equation of the circle with $AB$ as diameter is

In the co-axial system of circles $x^2 + y^2 + 2gx + c = 0$,where $g$ is a parameter,if $c > 0$,then the circles are

Let $C_1$ be the circle of radius $1$ with center at the origin. Let $C_2$ be the circle of radius $r$ with center at the point $A=(4,1)$,where $1 < r < 3$. Two distinct common tangents $PQ$ and $ST$ of $C_1$ and $C_2$ are drawn. The tangent $PQ$ touches $C_1$ at $P$ and $C_2$ at $Q$. The tangent $ST$ touches $C_1$ at $S$ and $C_2$ at $T$. Midpoints of the line segments $PQ$ and $ST$ are joined to form a line which meets the $x$-axis at a point $B$. If $AB=\sqrt{5}$,then the value of $r^2$ is

The equation of the circle having its centre on the line $x + 2y - 3 = 0$ and passing through the points of intersection of the circles ${x^2} + {y^2} - 2x - 4y + 1 = 0$ and ${x^2} + {y^2} - 4x - 2y + 4 = 0$ is:

The centres of all circles passing through the points of intersection of the circles $x^2+y^2+2x-2y+1=0$ and $x^2+y^2-2x+2y-2=0$ and having radius $\sqrt{14}$ lie on the curve

If $(p, q)$ is the centre of the circle which cuts the three circles $x^2+y^2-2x-4y+4=0$,$x^2+y^2+2x-4y+1=0$ and $x^2+y^2-4x-2y-11=0$ orthogonally,then $p+q=$