If the circle $x^2+y^2+4x-6y+c=0$ bisects the circumference of the circle $x^2+y^2-6x+4y-12=0$,then $c$ is equal to

If $\left(0, \frac{3}{4}\right)$ is the radical centre of the circles $S \equiv x^2+y^2+\alpha x+6y=0$,$S^{\prime} \equiv x^2+y^2+2\alpha x+\alpha y+6=0$ and $S^{\prime\prime} \equiv x^2+y^2+6\alpha x-\alpha y+3=0$,then the distance between the radical centre and the centre of the circle $S^{\prime}=0$ is:

The equation of the circle which passes through the origin and cuts orthogonally each of the circles $x^2+y^2-6x+8=0$ and $x^2+y^2-2x-2y-7=0$ is

The equation of the circle passing through the origin,having its center on the line $x + y = 4$,and intersecting the circle $x^2 + y^2 - 4x + 2y + 4 = 0$ orthogonally is:

The coordinates of the centre of the circle which bisects the circumferences of the circles $x^2 + y^2 = 1$,$x^2 + y^2 + 2x - 3 = 0$,and $x^2 + y^2 + 2y - 3 = 0$ are

If the lengths of tangents drawn from a point $P$ to the circles $x^2+y^2-8x+40=0$,$5x^2+5y^2-25x+80=0$,and $x^2+y^2-8x+16y+160=0$ are equal,then the point $P$ is: