If one of the diameters of the curve $x^{2}+y^{2}-4x-6y+9=0$ is a chord of a circle with centre $(1,1)$, the radius of this circle is

If the angle between the circles $x^2+y^2-4x-6y+k=0$ and $x^2+y^2+8x-4y+11=0$ is $\frac{\pi}{2}$,then the value of $k$ is

Find the equation of the circle whose diameter is the common chord of the circles $x^2 + y^2 - 8x + y - 15 = 0$ and $x^2 + y^2 - 4x + 4y - 42 = 0$.

The radius of the circle passing through the points of intersection of the circles $x^2+y^2+2x+4y+1=0$ and $x^2+y^2-2x-4y-4=0$ and intersecting the circle $x^2+y^2=6$ orthogonally is

The equation of the circle passing through the points of intersection of $x^2 + y^2 - 1 = 0$ and $x^2 + y^2 - 2x - 4y + 1 = 0$ and touching the line $x + 2y = 0$ is

$P$ is a point of intersection of the circles $S \equiv x^2+y^2-6x+2ky+1=0$ and $S' \equiv x^2+y^2+2kx-6y-7=0$. If the tangent at $P$ to $S=0$ passes through the centre of $S'=0$ and the tangent at $P$ to $S'=0$ passes through the centre of $S=0$,then the radius of $S'=0$ is