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:

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

The equation of the circle passing through $(1,1)$ and through the points of intersection of the circles $x^2+y^2+13x-3y=0$ and $2x^2+2y^2+4x-7y-25=0$ is

If $T_1 T_1^{\prime}$ and $T_2 T_2^{\prime}$ are the common tangents of the circles $S = x^2 + y^2 - 2x - 4y - 4 = 0$ and $S^{\prime} = x^2 + y^2 + 4x + 4y + 4 = 0$, where $T_1, T_1^{\prime}, T_2, T_2^{\prime}$ are the points of contact, then the distance between $T_1$ and $T_1^{\prime}$ is (in $\sqrt{6}$)

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