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

Let $C_i \equiv  x^2 + y^2 = i^2 (i = 1,2,3)$ are three circles. If there are $4i$ points on circumference of circle $C_i$. If no three of all the points on three circles are collinear then number of triangles which can be formed using these points whose circumcentre does not lie on origin, is-

The radical centre of the circles ${x^2} + {y^2} + 4x + 6y = 19,{x^2} + {y^2} = 9$ and ${x^2} + {y^2} - 2x - 2y = 5$ will be

If two circles ${(x - 1)^2} + {(y - 3)^2} = {r^2}$ and ${x^2} + {y^2} - 8x + 2y + 8 = 0$ intersect in two distinct points, 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 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 :