For all real values of $k$,the polar of the point $(2k, k-4)$ with respect to the circle $x^2+y^2-4x-6y+1=0$ passes through the point:

If $(4,2)$ and $(k,-3)$ are conjugate points with respect to $x^2+y^2-5x+8y+6=0$,then $k$ equals

The poles of the tangents to the circle $x^2+y^2=4$ with respect to the circle $(x+2)^2+y^2=8$ lie on

The condition for the lines $lx + my + n = 0$ and $l_1x + m_1y + n_1 = 0$ to be conjugate with respect to the circle $x^2 + y^2 = r^2$ is:

$A$ rectangle is formed by the lines $x=4, x=-2, y=5$,and $y=-2$. $A$ circle is drawn passing through the vertices of this rectangle. The pole of the line $y+2=0$ with respect to this circle is:

If the pole of the line $x+2by-5=0$ with respect to the circle $S \equiv x^2+y^2-4x-6y+4=0$ lies on the line $x+by+1=0$,then the polar of the point $(b,-b)$ with respect to the circle $S=0$ is