If two circles ${(x - 1)^2} + {(y - 3)^2} = {r^2}$ and ${x^2} + {y^2} - 8x + 2y + 8 = 0$ intersect in two distinct points, then

Figure shows $\Delta  ABC$ with $AB = 3, AC = 4$  &  $BC = 5$. Three circles $S_1, S_2$  &  $S_3$ have their centres on $A, B  $ &  $C$ respectively and they externally touches each other. The sum of areas of three circles 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 $(-8,1)$

Two tangents are drawn from a point $P$ on radical axis to the two circles touching at $Q$ and $R$ respectively then triangle formed by joining $PQR$ is

The equation of the circle which passes through the point of intersection of circles ${x^2} + {y^2} - 8x - 2y + 7 = 0$ and ${x^2} + {y^2} - 4x + 10y + 8 = 0$ and having its centre on $y$ - axis, will be

Suppose we have two circles of radius 2 each in the plane such that the distance between their centers is $2 \sqrt{3}$. The area of the region common to both circles lies between