A rhombus is inscribed in the region common t | vedclass

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(A) The given circles are $C_1: x^2 + y^2 - 4x - 12 = 0$ with center $(2, 0)$ and radius $r = \sqrt{2^2 + 0^2 - (-12)} = \sqrt{16} = 4$.The second cir

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$8\sqrt{3} 	ext{ sq. units}$

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(A) The given circles are $C_1: x^2 + y^2 - 4x - 12 = 0$ with center $(2, 0)$ and radius $r = \sqrt{2^2 + 0^2 - (-12)} = \sqrt{16} = 4$.The second circle is $C_2: x^2 + y^2 + 4x - 12 = 0$ with center $(-2, 0)$ and radius $r = 4$.The common chord is obtained by subtracting the two equations: $(x^2 + y^2 - 4x - 12) - (x^2 + y^2 + 4x - 12) = 0$,which gives $-8x = 0$,so $x = 0$ (the $y$-axis).The intersection points of the circles with the $y$-axis are found by setting $x = 0$ in either equation: $y^2 - 12 = 0$,so $y = \pm 2\sqrt{3}$.The vertices of the rhombus are $(2, 0)$,$(-2, 0)$,$(0, 2\sqrt{3})$,and $(0, -2\sqrt{3})$.The diagonals of the rhombus are $d_1 = 2 - (-2) = 4$ and $d_2 = 2\sqrt{3} - (-2\sqrt{3}) = 4\sqrt{3}$.The area of the rhombus is $\frac{1}{2} 	imes d_1 	imes d_2 = \frac{1}{2} 	imes 4 	imes 4\sqrt{3} = 8\sqrt{3} 	ext{ sq. units}$.

$A$ 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 centers of the circles. The area of the rhombus is

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