The equation of the circle which passes through the point $(1, 1)$ and intersects the given circles $x^2 + y^2 + 2x + 4y + 6 = 0$ and $x^2 + y^2 + 4x + 6y + 2 = 0$ orthogonally is:

If the equation of the circle which passes through the point $(1,1)$ and cuts both the circles $x^2+y^2-4x-6y+4=0$ and $x^2+y^2+6x-4y+15=0$ orthogonally is $x^2+y^2+2gx+2fy+c=0$,then $5g+2f+c=$

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 $\sqrt{6}$)

The number of common tangents that can be drawn to the circles $x^2+y^2-2x-2y-23=0$ and $x^2+y^2-4x-4y-1=0$ is

If $x^2+y^2-a^2+\lambda(x \cos \alpha+y \sin \alpha-p)=0$ is the smallest circle through the points of intersection of $x^2+y^2=a^2$ and $x \cos \alpha+y \sin \alpha=p$,where $0 < p < a$,then $\lambda=$

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: