If the inverse point of $(1,1)$ with respect to the circle $x^2+y^2-4x-6y+12=0$ is $(h, k)$,then $h+k$ is equal to

$A$ 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:

The pole of the line $2x + 3y = 4$ with respect to the circle $x^2 + y^2 = 64$ is:

The polar of the origin $(0, 0)$ with respect to the circle ${x^2} + {y^2} + 2\lambda x + 2\mu y + c = 0$ touches the circle ${x^2} + {y^2} = {r^2}$,if

The pole of the straight line $9x + y - 28 = 0$ with respect to the circle $2x^2 + 2y^2 - 3x + 5y - 7 = 0$ 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