A rhombus is inscribed in the region common t | vedclass

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(A) The given circles are $C_1: x^2 + y^2 - 4x - 12 = 0$ (centre $(2, 0)$,radius $r = \sqrt{2^2 + 0^2 + 12} = 4$) and $C_2: x^2 + y^2 + 4x - 12 = 0$ (

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$8\sqrt{3}$ sq.units

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(A) The given circles are $C_1: x^2 + y^2 - 4x - 12 = 0$ (centre $(2, 0)$,radius $r = \sqrt{2^2 + 0^2 + 12} = 4$) and $C_2: x^2 + y^2 + 4x - 12 = 0$ (centre $(-2, 0)$,radius $r = 4$).The line joining the centres is the $x$-axis $(y = 0)$.The common chord is obtained by subtracting the equations: $(x^2 + y^2 - 4x - 12) - (x^2 + y^2 + 4x - 12) = 0$ $\Rightarrow -8x = 0$ $\Rightarrow x = 0$ (the $y$-axis).The vertices of the rhombus on the $x$-axis are the centres of the circles,$(-2, 0)$ and $(2, 0)$,so the length of one diagonal is $d_1 = 2 - (-2) = 4$.The other diagonal lies on the common chord $x = 0$. The intersection points of $x = 0$ with $x^2 + y^2 - 4x - 12 = 0$ are $y^2 - 12 = 0 \Rightarrow y = \pm 2\sqrt{3}$.Thus,the length of the second diagonal is $d_2 = 2\sqrt{3} - (-2\sqrt{3}) = 4\sqrt{3}$.The area of the rhombus is $A = \frac{1}{2} 	imes d_1 	imes d_2 = \frac{1}{2} 	imes 4 	imes 4\sqrt{3} = 8\sqrt{3}$ sq.units.

$A$ rhombus is inscribed in the region common to the two circles $x^2 + y^2 - 4x - 12 = 0$ and $x^2 + y^2 + 4x - 12 = 0$,with two of its vertices on the line joining the centres of the circles. The area of the rhombus is:

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