The point of concurrence of all conjugate lines of the line $5x + 7y - 78 = 0$ with respect to the circle $x^2 + y^2 + 6x + 8y - 96 = 0$ is

The normal to a circle $S=0$ at $P(1,3)$ is $x+2y=7$ and it has another normal at $Q(3,5)$ which is the polar of the point $A(7, -1/2)$ with respect to the circle $x^2+y^2-4x+6y-12=0$. Then,the equation of the circle $S=0$ 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 straight line $9x + y - 28 = 0$ with respect to the circle $2x^2 + 2y^2 - 3x + 5y - 7 = 0$ is

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

If $A(2, c)$ and $B(d, 2)$ are two points such that the polar of one point with respect to the circle $x^2+y^2=16$ passes through the other,then $c+d=$