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

8 \sqrt{3}

sq. units

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4 \sqrt{3}

sq. units

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6 \sqrt{3}

sq. units

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None

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Solution

The correct option is A $8 \sqrt{3}$ sq. units
Circles with centre $(2, 0)$ and $(- 2, 0)$ each with radius $4.$
$\Rightarrow$ y -axis is their common chord.
$Δ$ ABC is equilateral. Hence, are of ADBC is $\frac{2. \sqrt{3}}{4} (4)^{2}$
$= 8 \sqrt{3}$ .

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Similar questions

x^{2} + y^{2} - 4 x - 12 = 0

x^{2} + y^{2} + 4 x - 12 = 0

y^{2} = 4 x

x^{2} = 4 y

\frac{16}{3}

y^{2} = 4 a x

x^{2} = 4 a y

\frac{16 a^{2}}{3}

x^{2} + 2 y - 1 = 0, y^{2} + 4 x - 4 = 0 and y^{2} - 4 x - 4 = 0,

The area (in sq. units) of the region ${(x, y) : y^{2} \geq 2 x a n d x^{2} + y^{2} \leq 4 x, x \geq 0, y \geq 0}$ is :

x^{2} + y^{2} + 10 x + 12 y + c = 0

27 \sqrt{3}

c

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