A rhombus is inscribed in the region common t | vedclass

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(D) The given circles are $C_1: x^2+y^2-4x-12=0$ and $C_2: x^2+y^2+4x-12=0$.Their centers are $A(-2, 0)$ and $B(2, 0)$, and both have radius $r = \sqr

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$8$

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(D) The given circles are $C_1: x^2+y^2-4x-12=0$ and $C_2: x^2+y^2+4x-12=0$.Their centers are $A(-2, 0)$ and $B(2, 0)$, and both have radius $r = \sqrt{2^2+0^2+12} = 4$.The common chord is obtained by subtracting the equations: $(x^2+y^2+4x-12) - (x^2+y^2-4x-12) = 0$, which gives $8x = 0$, or $x = 0$ (the $y$-axis).The intersection points $C$ and $D$ are found by substituting $x=0$ into $x^2+y^2-4x-12=0$, giving $y^2 = 12$, so $y = \pm 2\sqrt{3}$. Thus, $C(0, 2\sqrt{3})$ and $D(0, -2\sqrt{3})$.The diagonals of the rhombus are $AB$ (length $d_1 = 4$) and $CD$ (length $d_2 = 4\sqrt{3}$).The area of the rhombus is $\frac{1}{2} 	imes d_1 	imes d_2 = \frac{1}{2} 	imes 4 	imes 4\sqrt{3} = 8\sqrt{3}$ square units.

$A$ rhombus is inscribed in the region common to the two circles $x^2+y^2-4x-12=0$ and $x^2+y^2+4x-12=0$. If the line joining the centres of these circles and the common chord of them are the diagonals of this rhombus, then the area (in square units) of the rhombus is (in $\sqrt{3}$)

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