A variable line $ax + by + c = 0$, where $a, b, c$ are in $A.P.$, is normal to a circle $(x - \alpha)^2 + (y - \beta)^2 = \gamma$ , which is orthogonal to circle $x^2 + y^2- 4x- 4y-1 = 0$. The value of $\alpha + \beta + \gamma$ is equal to

Two circles ${S_1} = {x^2} + {y^2} + 2{g_1}x + 2{f_1}y + {c_1} = 0$ and ${S_2} = {x^2} + {y^2} + 2{g_2}x + 2{f_2}y + {c_2} = 0$ cut each other orthogonally, then

The line $L$ passes through the points of intersection of the circles ${x^2} + {y^2} = 25$ and ${x^2} + {y^2} - 8x + 7 = 0$. The length of perpendicular from centre of second circle onto the line $L$, is

If circles ${x^2} + {y^2} + 2ax + c = 0$and ${x^2} + {y^2} + 2by + c = 0$ touch each other, then 

If the circles ${x^2} + {y^2} = 4,{x^2} + {y^2} - 10x + \lambda = 0$ touch externally, then $\lambda $ is equal to 

The equation of the circle which passes through the intersection of ${x^2} + {y^2} + 13x - 3y = 0$and $2{x^2} + 2{y^2} + 4x - 7y - 25 = 0$ and whose centre lies on $13x + 30y = 0$ is