Equation of radical axis of the circles ${x^2} + {y^2} - 3x - 4y + 5 = 0$, $2{x^2} + 2{y^2} - 10x$$ - 12y + 12 = 0$ is

The equation of the circle which passing through the point $(2a,\,0)$ and whose radical axis is $x = \frac{a}{2}$ with respect to the circle ${x^2} + {y^2} = {a^2},$ will be 

Two given circles ${x^2} + {y^2} + ax + by + c = 0$ and ${x^2} + {y^2} + dx + ey + f = 0$ will intersect each other orthogonally, only when

The circles ${x^2} + {y^2} + 4x + 6y + 3 = 0$ and $2({x^2} + {y^2}) + 6x + 4y + C = 0$ will cut orthogonally, if $C$ equals

If the circles $(x+1)^2+(y+2)^2=r^2$ and $x^2+y^2-4 x-4 y+4=0$ intersect at exactly two distinct points, then

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-