If $x^2+y^2-6x-8y+12=0$ and $x^2+y^2-4x+6y+k=0$ cut orthogonally,then $k=$

The equation of the circle passing through the points of intersection of the two orthogonal circles $S_1 = x^2 + y^2 + kx - 4y - 1 = 0$ and $S_2 = 3x^2 + 3y^2 - 14x + 23y - 15 = 0$ and passing through the point $(-1, -1)$ is:

The equation of the circle passing through the points of intersection of the circles $x^2+y^2+6x+4y-12=0$ and $x^2+y^2-4x-6y-12=0$ and having radius $\sqrt{13}$ is

The equation of the circle described on the chord $3x + y + 5 = 0$ of the circle $x^2 + y^2 = 16$ as diameter 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 centre $(\alpha, \beta)$ of a circle cutting the circles $x^2+y^2-2y-3=0$ and $x^2+y^2+4x+3=0$ orthogonally lies on the line $2x-3y+4=0$,then $2\alpha+\beta=$