The radical axis of the circles $x^2+y^2+5x+4y-5=0$ and $x^2+y^2-3x+5y-6=0$ is:

Two circles $S_1 = x^2 + y^2 + 2g_1x + 2f_1y + c_1 = 0$ and $S_2 = x^2 + y^2 + 2g_2x + 2f_2y + c_2 = 0$ cut each other orthogonally,then:

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 described on the chord $3x + y + 5 = 0$ of the circle $x^2 + y^2 = 16$ as diameter is:

The point $(3, -4)$ lies on both the circles $x^2 + y^2 - 2x + 8y + 13 = 0$ and $x^2 + y^2 - 4x + 6y + 11 = 0$. Then,the angle between the circles is

If a diameter of the circle $x^2+y^2-4x+6y-12=0$ is a chord of a circle $S$ whose centre is at $(-3, 2)$,then the radius of $S$ is