Let a circle $C_1 \equiv  x^2 + y^2 - 4x + 6y + 1 = 0$ and circle $C_2$ is such that it's centre is image of centre of $C_1$ about $x-$axis and radius of $C_2$ is equal to radius of $C_1$, then area of $C_1$ which is not common with $C_2$ is -

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

If the two circles, $x^2 + y^2 + 2 g_1x + 2 f_1y = 0\, \& \,x^2 + y^2 + 2 g_2x + 2 f_2y = 0$ touch each then:

The number of circles touching the line $y - x = 0$ and the $y$-axis is

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