The equation of circle passing through the points of intersection of circles ${x^2} + {y^2} - 6x + 8 = 0$ and ${x^2} + {y^2} = 6$ and point $(1, 1)$, is

The equation of the circle which intersects circles ${x^2} + {y^2} + x + 2y + 3 = 0$, ${x^2} + {y^2} + 2x + 4y + 5 = 0$and ${x^2} + {y^2} - 7x - 8y - 9 = 0$ at right angle, will be

The common tangent to the circles $x^2 + y^2 = 4$ and $x^2 + y^2 + 6x + 8y - 24 = 0$ also passes through the point

The two circles ${x^2} + {y^2} - 2x - 3 = 0$ and ${x^2} + {y^2} - 4x - 6y - 8 = 0$ are such that

If the line $x\, cos \theta + y\, sin \theta = 2$ is the equation of a transverse common tangent to the circles $x^2 + y^2 = 4$ and $x^2 + y^2 - 6 \sqrt{3} \,x - 6y + 20 = 0$, then the value of $\theta$ is :

Choose the correct statement about two circles whose equations are given below

$x^{2}+y^{2}-10 x-10 y+41=0$

$x^{2}+y^{2}-22 x-10 y+137=0$