The equation of the circle passing through the point $(-2, 4)$ and through the points of intersection of the circle ${x^2} + {y^2} - 2x - 6y + 6 = 0$ and the line $3x + 2y - 5 = 0$, is

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

The tangent to the circle $C_1 : x^2 + y^2 - 2x- 1\, = 0$ at the point $(2, 1)$ cuts off a chord of length $4$ from a circle $C_2$ whose centre is $(3, - 2)$. The radius of $C_2$ 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

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

Let the latus ractum of the parabola $y ^{2}=4 x$ be the common chord to the circles $C _{1}$ and $C _{2}$ each of them having radius $2 \sqrt{5}$. Then, the distance between the centres of the circles $C _{1}$ and $C _{2}$ is