A rhombus is inscribed in the region common to the two circles x2+y2-4 x-12=0 and x2+y2+4 x-12=0 with two of its vertices on the line joining the centers of the circles. The area of rhombus is

Question

A rhombus is inscribed in the region common to the two circles x2+y2-4 x-12=0 and x2+y2+4 x-12=0 with two of its vertices on the line joining the centers of the circles. The area of rhombus is (A) 8 √3 (B) 4 √2 (C) 16 √3 (D) 16 √2

Tardigrade Admin · Accepted Answer

Circles with centers A(-2,0) and B(+2,0), each of radius 4 has Y-axis as their common chord. <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/3aa57c7afd138bc79c9fe5c78412a3a0-.png" /> ⇒ Δ A B C is an equilateral triangle. Hence, the area of rhombus A D B C is =2 ⋅ (√3/4) × 42=8 √3

Q. A rhombus is inscribed in the region common to 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 rhombus is

$83$

$42$

$163$

$162$

Circles with centers $A (- 2, 0)$ and $B (+ 2, 0)$ , each of radius $4$ has $Y$ -axis as their common chord.

$\Rightarrow Δ A BC$ is an equilateral triangle.
Hence, the area of rhombus $A D BC$ is
$= 2 \cdot \frac{3}{4} \times 4^{2} = 83$

Q. A rhombus is inscribed in the region common to the two circles x2+y2−4x−12=0 and x2+y2+4x−12=0 with two of its vertices on the line joining the centers of the circles. The area of rhombus is

83​

42​

163​

162​