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:

Let $S \equiv x^2+y^2-8x+10y+5=0$ be a circle. Let $P(1,1)$ and $Q(1,-1)$ be two points. Then the point of intersection of the polar of $P$ with respect to $S=0$ and the chord with $Q$ as mid-point to $S=0$ 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 line $\frac{x}{a} + \frac{y}{b} = 1$ with respect to the circle $x^2 + y^2 = c^2$ is

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

If the polar of a point on the circle $x^2+y^2=p^2$ with respect to the circle $x^2+y^2=q^2$ touches the circle $x^2+y^2=r^2$,then $p, q, r$ are in