The equation of the circle through the point of intersection of the circles ${x^2} + {y^2} – 8x – 2y + 7 = 0$, ${x^2} + {y^2} – 4x + 10y + 8 = 0$ and $(3, -3)$ is

Circles ${(x + a)^2} + {(y + b)^2} = {a^2}$ and ${(x + \alpha )^2}$ $ + {(y + \beta )^2} = $ ${\beta ^2}$ cut orthogonally, if

The gradient of the radical axis of the circles ${x^2} + {y^2} – 3x – 4y + 5 = 0$ and $3{x^2} + 3{y^2} – 7x + 8y + 11 = 0$ is

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

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