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

The condition of the curves $a{x^2} + b{y^2} = 1$and $a'{x^2} + b'{y^2} = 1$ to intersect each other orthogonally, is

The radical axis of two circles and the line joining their centres are

The circles $x^2 + y^2 + 2x -2y + 1 = 0$ and $x^2 + y^2 -2x -2y + 1 = 0$ touch each  other :-

The minimum distance between any two points $P _{1}$ and $P _{2}$ while considering point $P _{1}$ on one circle and point $P _{2}$ on the other circle for the given circles' equations

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

$x^{2}+y^{2}-24 x-10 y+160=0$ is .........

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-