Let $C_1$ and $C_2$ be the centres of the circles $x^2 + y^2 -2x -2y -2 = 0$ and $x^2 + y^2 - 6x-6y + 14 = 0$ respectively. If $P$ and $Q$ are the points of intersection of these circles, then the area (in sq. units) of the quadrilateral $PC_1QC_2$ is ............. $\mathrm{sq. \, units}$

Two circles of radii $4$ cms $\&\,\, 1\,\, cm$ touch each other externally and $\theta$ is the angle contained by their direct common tangents. Then $sin \theta =$

If ${x^2} + {y^2} + px + 3y - 5 = 0$ and ${x^2} + {y^2} + 5x$ $ + py + 7 = 0$ cut orthogonally, then $p$ 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 circle passing through the intersection of the circles, $x^{2}+y^{2}-6 x=0$ and $x^{2}+y^{2}-4 y=0$ having its centre on the line, $2 x-3 y+12=0$, also passes through the point

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