The point of concurrence of the polars of the variable point $(2t, t-4)$,where $t \in R$,with respect to the circle $x^2+y^2-4x-6y+1=0$ is

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

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

An ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with eccentricity $e=\frac{2 \sqrt{2}}{3}$ is inscribed in a circle $x^2+y^2=18$ such that the length of its major axis is equal to the diameter of this circle. The locus of the poles of all the tangents of the circle with respect to the ellipse is

Consider the point $P(\alpha, \beta)$ on the line $2x+y=1$. If $P$ and $(3,2)$ are conjugate points with respect to the circle $x^2+y^2=4$,then $\alpha+\beta=$

The inverse point of $(1, 2)$ with respect to the circle $x^2 + y^2 - 4x - 6y + 9 = 0$ is