The equation of the circle passing through the points of intersection of two circles $x^2+y^2+2x+3y+1=0$ and $x^2+y^2+4x+3y+2=0$ and the point $(-1,1)$ is

If the circle $x^2+y^2-6x-12y+1=0$ cuts another circle $C$ orthogonally and the centre of the circle $C$ is $(-4, 2)$,then its radius is

$(a, 0)$ and $(b, 0)$ are centers of two circles belonging to a coaxial system of which the $y$-axis is the radical axis. If the radius of one of the circles is $r$,then the radius of the other circle is:

Suppose $A(2,3)$ and $B$ are the points of intersection of two circles. The points $P$ lying on one circle and $Q$ lying on the other circle are such that $BP$ and $BQ$ constitute the diameters of the circles. If the slopes of the radical axis and $PQ$ are $3/4$ and $a/b$ respectively,then the value of $3a+4b$ is

The equation of the circle touching the line $2x+3y+1=0$ at the point $(1,-1)$ and orthogonal to the circle which has the line segment having end points $(0,-1)$ and $(-2,3)$ as diameter,is

If the circles $x^2+y^2+kx+4y+2=0$ and $2(x^2+y^2)-4x-3y+k=0$ cut orthogonally,then $k=$