The radical centre of the circles $x^2+y^2=1$,$x^2+y^2-2x-3=0$ and $x^2+y^2-2y-3=0$ is

One of the limit points of the coaxial system of circles containing $x^2 + y^2 - 6x - 6y + 4 = 0$ and $x^2 + y^2 - 2x - 4y + 3 = 0$ is:

If $(a, b)$ is the common point for the circles $x^2+y^2-4x+4y-1=0$ and $x^2+y^2+2x-4y+1=0$,then $a^2+b^2=$

The equation of a circle that intersects the circle $x^2 + y^2 + 14x + 6y + 2 = 0$ orthogonally and whose centre is $(0, 2)$ is

If the angle between the circles $x^2+y^2+2x-4y+1=0$ and $x^2+y^2-4x-2y+c=0$ is $\frac{\pi}{4}$,then $c=$

The centres of all circles passing through the points of intersection of the circles $x^2+y^2+2x-2y+1=0$ and $x^2+y^2-2x+2y-2=0$ and having radius $\sqrt{14}$ lie on the curve