Question

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 Sq. units) of the rhombus is:

• A. \( 16\sqrt{3} \)
• B. \( 4\sqrt{3} \)
• C. \( 12\sqrt{3} \)
• D. \( 8\sqrt{3} \)
  \

Hint:

For problems involving circles and common chords, use the standard form of the equation to determine the centre and radius. The common chord length can be found by solving for intersection points.

Answer 1

The Correct Option is D

Step 1: Find the centres and radii of the circles
The general equation of a circle is: \[ x^2 + y^2 + 2gx + 2fy + c = 0. \] For the first circle: \[ x^2 + y^2 - 4x - 12 = 0. \] Rewriting, \[ (x - 2)^2 + y^2 = 16. \] Thus, the centre is \( (2, 0) \) and the radius is \( r = 4 \). For the second circle: \[ x^2 + y^2 + 4x - 12 = 0. \] Rewriting, \[ (x + 2)^2 + y^2 = 16. \] Thus, the centre is \( (-2, 0) \) and the radius is \( r = 4 \). Step 2: Find the length of the line joining the centres
The distance between the centres \( (2,0) \) and \( (-2,0) \) is: \[ d = \sqrt{(2 - (-2))^2 + (0 - 0)^2} = \sqrt{(4)^2} = 4. \] Step 3: Find the common chord
The equation of the common chord is obtained by subtracting the two circle equations: \[ (x^2 + y^2 - 4x - 12) - (x^2 + y^2 + 4x - 12) = 0. \] \[ -4x - 4x = 0 \Rightarrow -8x = 0. \] \[ x = 0. \] This represents the common chord along the \( y \)-axis. To find its length, we substitute \( x = 0 \) in one of the circles: \[ y^2 = 12. \] \[ y = \pm 2\sqrt{3}. \] So, the total length of the common chord is: \[ 2(2\sqrt{3}) = 4\sqrt{3}. \] Step 4: Compute the area of the rhombus
The diagonals of the rhombus are: \[ d_1 = 4, \quad d_2 = 4\sqrt{3}. \] The area of a rhombus is given by: \[ A = \frac{1}{2} d_1 d_2. \] \[ A = \frac{1}{2} (4 \times 4\sqrt{3}). \] \[ A = \frac{16\sqrt{3}}{2} = 8\sqrt{3}. \] Step 5: Conclusion
Thus, the correct answer is: \[ \mathbf{8\sqrt{3}} \] \bigskip