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 $a^2+b^2=$

Find the equation of the circle whose diameter is the common chord of the circles $x^2 + y^2 - 8x + y - 15 = 0$ and $x^2 + y^2 - 4x + 4y - 42 = 0$.

The limiting points of the co-axial system containing the two circles $x^2+y^2+2x-2y+2=0$ and $25(x^2+y^2)-10x-80y+65=0$ are

$(a, 0)$ and $(b, 0)$ are the centres of two circles belonging to a coaxial system of which the $y$-axis is the radical axis. If the radius of one of the circles is $r$,then the radius of the other circle is

Suppose the circle $S: x^2+y^2+2gx+2fy+c=0$ cuts orthogonally the two circles $S': x^2+y^2-4x-6y+11=0$ and $S'': x^2+y^2-10x-4y+21=0$. If the centre of $S=0$ lies on the bisector of the angle between the positive coordinate axes,then $2g+2f+c=$

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 $c=$