The equation of a circle that intersects the circle ${x^2} + {y^2} + 14x + 6y + 2 = 0$orthogonally and whose centre is $(0, 2)$ is

For the two circles $x^2 + y^2 = 16$ and $x^2 + y^2 -2y = 0,$ there is/are

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

Coordinates of the centre of the circle which bisects the circumferences of the circles

$x^2 + y^2 = 1 ; x^2 + y^2 + 2x - 3 = 0$ and $x^2 + y^2 + 2y - 3 = 0$ 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:

Let $C$ be a circle passing through the points $A (2,-1)$ and $B (3,4)$. The line segment $AB$ is not a diameter of $C$. If $r$ is the radius of $C$ and its centre lies on the circle $(x-5)^{2}+(y-1)^{2}=\frac{13}{2}$, then $r^{2}$ is equal to