The equation of the line perpendicular to the radical axis of two circles $x^2+y^2-5x+6y+12=0$ and $x^2+y^2+6x-4y-14=0$,and passing through $(1,1)$ is:

The point at which the circles $x^2+y^2-4x-4y+7=0$ and $x^2+y^2-12x-10y+45=0$ touch each other is:

If $d$ is the distance between the centres of two circles,$r_1$ and $r_2$ are their radii,and $d = r_1 + r_2$,then

The circle $S=0$ cuts the circles $C_1=x^2+y^2-8x-2y+16=0$ and $C_2=x^2+y^2-4x-4y-1=0$ orthogonally. If the common chord of $S=0$ and $C_1=0$ is $2x+13y-15=0$,then the centre of $S=0$ is

Let the latus rectum of the parabola $y^{2} = 4x$ be the common chord to the circles $C_{1}$ and $C_{2}$,each of them having radius $2\sqrt{5}$. Then,the distance between the centres of the circles $C_{1}$ and $C_{2}$ is

If $y = 2x$ is a chord of the circle $x^2 + y^2 - 10x = 0$,then the equation of the circle of which this chord is a diameter is: