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

Let the mirror image of a circle $c_{1}: x^{2}+y^{2}-2 x-$ $6 y+\alpha=0$ in line $y=x+1$ be $c_{2}: 5 x^{2}+5 y^{2}+10 g x$ $+10 f y +38=0$. If $r$ is the radius of circle $c _{2}$, then $\alpha+6 r^{2}$ is equal to$.....$

Circles ${x^2} + {y^2} + 2gx + 2fy = 0$ and ${x^2} + {y^2}$ $ + 2g'x + 2f'y = $ $0$ touch externally, if

Suppose we have two circles of radius 2 each in the plane such that the distance between their centers is $2 \sqrt{3}$. The area of the region common to both circles lies between

The circles ${x^2} + {y^2} - 10x + 16 = 0$ and ${x^2} + {y^2} = {r^2}$ intersect each other in two distinct points, if

Radius of circle touching $y-$axis at point $P(0,2)$ and circle $x^2 + y^2 = 16$ internally-