Let C_1 and C_2 be the centres of the circles | vedclass

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(D) For the circle $x^2 + y^2 - 2x - 2y - 2 = 0$,the centre $C_1 = (1, 1)$ and radius $r_1 = \sqrt{1^2 + 1^2 - (-2)} = \sqrt{4} = 2$.For the circle $x

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$4$

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(D) For the circle $x^2 + y^2 - 2x - 2y - 2 = 0$,the centre $C_1 = (1, 1)$ and radius $r_1 = \sqrt{1^2 + 1^2 - (-2)} = \sqrt{4} = 2$.For the circle $x^2 + y^2 - 6x - 6y + 14 = 0$,the centre $C_2 = (3, 3)$ and radius $r_2 = \sqrt{3^2 + 3^2 - 14} = \sqrt{18 - 14} = \sqrt{4} = 2$.The distance between the centres $C_1C_2 = \sqrt{(3-1)^2 + (3-1)^2} = \sqrt{2^2 + 2^2} = \sqrt{8} = 2\sqrt{2}$.In the quadrilateral $PC_1QC_2$,the sides are $C_1P = C_1Q = r_1 = 2$ and $C_2P = C_2Q = r_2 = 2$.Since all sides are equal to $2$,the quadrilateral is a rhombus.The diagonals are $C_1C_2 = 2\sqrt{2}$ and $PQ$. The length of the common chord $PQ$ can be found using the formula $PQ = \frac{2r_1r_2}{d} \sin(	heta)$,or by noting that in $	riangle PC_1C_2$,$C_1P=2, C_2P=2, C_1C_2=2\sqrt{2}$. Since $2^2 + 2^2 = (2\sqrt{2})^2$,$	riangle PC_1C_2$ is a right-angled triangle at $P$.The area of $	riangle PC_1C_2 = \frac{1}{2} 	imes 2 	imes 2 = 2$.The area of the quadrilateral $PC_1QC_2 = 2 	imes (	ext{Area of } 	riangle PC_1C_2) = 2 	imes 2 = 4$ sq. units.

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 ............. $sq. \, units$.

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