The equation of the line joining the centres of the circles belonging to the coaxial system of circles $4x^2 + 4y^2 - 12x + 6y - 3 + \lambda(x + 2y - 6) = 0$ is

The equation of the circle passing through $(1,2)$ and the points of intersection of the circles $x^2+y^2-8x-6y+21=0$ and $x^2+y^2-2x-15=0$ is:

$A$ circle $S$ given by $x^2+y^2-14x+6y+33=0$ cuts the $X$-axis at $A$ and $B$ $(OB > OA)$. $C$ is the midpoint of $AB$. $L$ is a line through $C$ with slope $-1$. If $L$ is the diameter of a circle $S^{\prime}$ and also the radical axis of the circles $S$ and $S^{\prime}$,then the equation of the circle $S^{\prime}$ 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

Two tangents are drawn from a point $P$ on the radical axis to two circles,touching them at $Q$ and $R$ respectively. Then,the triangle formed by joining $P, Q,$ and $R$ is:

If the circles $x^2+y^2+2kx-4y+1=0$ and $x^2+y^2-8x-12y+43=0$ touch each other,then $k=$