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

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

A circle $S$ passes through the point $(0,1)$ and is orthogonal to the circles $(x-1)^2+y^2=16$ and $x^2+y^2=1$. Then

$(A)$ radius of $S$ is $8$

$(B)$ radius of $S$ is $7$

$(C)$ centre of $S$ is $(-7,1)$

$(D)$ centre of $S$ is $(-8,1)$

Radius of circle touching $y-$axis at point $P(0,2)$ and circle $x^2 + y^2 = 16$ internally-

Let the centre of a circle, passing through the point $(0,0),(1,0)$ and touching the circle $x^2+y^2=9$, be $(h, k)$. Then for all possible values of the coordinates of the centre $(h, k), 4\left(h^2+k^2\right)$ is equal to .............

The equation of the circle which intersects circles ${x^2} + {y^2} + x + 2y + 3 = 0$, ${x^2} + {y^2} + 2x + 4y + 5 = 0$and ${x^2} + {y^2} - 7x - 8y - 9 = 0$ at right angle, will be