If the circles ${x^2} + {y^2} + 2x + 2ky + 6 = 0$ and ${x^2} + {y^2} + 2ky + k = 0$ intersect orthogonally, then $k$ is

$P, Q$ and $R$ are the centres and ${r_1},\,\,{r_2},\,\,{r_3}$ are the radii respectively of three co-axial circles, then $QRr_1^2 + RP\,r_2^2 + PQr_3^2$ is equal to

The equation of the circle which passes through the intersection of ${x^2} + {y^2} + 13x - 3y = 0$and $2{x^2} + 2{y^2} + 4x - 7y - 25 = 0$ and whose centre lies on $13x + 30y = 0$ is

If $P$ and $Q$ are the points of intersection of the circles ${x^2} + {y^2} + 3x + 7y + 2p - 5 = 0$ and ${x^2} + {y^2} + 2x + 2y - {p^2} = 0$ then there is a circle passing through $P, Q$ and $(1, 1)$ for:

If a circle passes through the point $(a , b) \&$ cuts the circle $x^2 + y^2= K^2$ orthogonally, then the equation of the locus of its centre is :

The equation of the circle through the points of intersection of ${x^2} + {y^2} - 1 = 0$, ${x^2} + {y^2} - 2x - 4y + 1 = 0$ and touching the line $x + 2y = 0$, is