The equation of the radical axis of the circles $x^2+y^2+4x+6y+7=0$ and $4x^2+4y^2+8x+12y-9=0$ is:

If a circle passing through the point $(1,1)$ cuts the circles $x^2+y^2+4x-5=0$ and $x^2+y^2-4y+3=0$ orthogonally,then the center of that circle is:

The equation of the circle whose diameter is the common chord of the circles $x^2+y^2+2x+3y+1=0$ and $x^2+y^2+4x+3y+2=0$ is

Let the centre of a circle,passing through the points $(0,0)$ and $(1,0)$ and touching the circle $x^2+y^2=9$,be $(h, k)$. Then for all possible values of the coordinates of the centre $(h, k)$,$4(h^2+k^2)$ is equal to .............

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

The circle ${x^2 + y^2 + 2gx + 2fy + c = 0}$ bisects the circumference of the circle ${x^2 + y^2 + 2g'x + 2f'y + c' = 0}$,if