Let and be the centres of the circles and respectively. If and are the points of intersection of these circles, then the area (in sq units) of the quadrilateral is
- A.8
- B.4
- C.6
- D.9
Show answer and solution
Correct option: B
Solution
The centres are C₁ = (1,1) with radius r₁ = √4 = 2, and C₂ = (3,3) with radius r₂ = √4 = 2. The distance C₁C₂ = √8 = 2√2. Since both radii equal 2 and C₁C₂ = 2√2, triangle PC₁C₂ is a right isosceles triangle (2² + 2² = 8 = (2√2)²), so the angle at P is 90°. The quadrilateral PC₁QC₂ is a rhombus with diagonals PQ and C₁C₂. The common chord PQ is perpendicular to C₁C₂; its half-length = √(r₁² - (C₁C₂/2)²) = √(4 - 2) = √2, so PQ = 2√2. Area of rhombus = (1/2)×d₁×d₂ = (1/2)×2√2×2√2 = (1/2)×8 = 4.
[JEE Main 2019 · 12 Jan · Shift I]