<p>If a circle of radius 2 unit touches the \(y\)-axis at the origin, \('O'\) and intersects the lines \(y = (2-\sqrt{3})x\) and \(y = -(2+\sqrt{3})x\) in the I and IV quadrants at \(A\) and \(B\) respectively, then area of \(\triangle AOB\) (in square units) is:</p>
Step-by-Step Solution
Key Concept: Since the circle touches the y-axis at origin with radius 2, its center is at (2,0). Use the perpendicular distance from center to each line to find the chord lengths, then apply the distance formula for triangle area using the intersection points.
<p><strong>Step 1: Identify the circle equation</strong></p><p>Circle touches y-axis at O(0,0) with radius 2 → center at (2,0)<br>Circle equation: (x-2)² + y² = 4</p><p><strong>Step 2: Find intersection point A with line y = (2-√3)x</strong></p><p>Substitute into circle: (x-2)² + (2-√3)²x² = 4<br>(x-2)² + (7-4√3)x² = 4<br>x² - 4x + 4 + (7-4√3)x² = 4<br>(8-4√3)x² - 4x = 0<br>x(2-√3)x - x = 0<br>x = 0 or x = 2/(2-√3) = 2(2+√3)</p><p>For x = 2(2+√3): y = (2-√3)·2(2+√3) = 2(4-3) = 2<br>So A = (2(2+√3), 2)</p><p><strong>Step 3: Find intersection point B with line y = -(2+√3)x</strong></p><p>Substitute: (x-2)² + (2+√3)²x² = 4<br>(x-2)² + (7+4√3)x² = 4<br>(8+4√3)x² - 4x = 0<br>x = 0 or x = 2/(2+√3) = 2(2-√3)</p><p>For x = 2(2-√3): y = -(2+√3)·2(2-√3) = -2(4-3) = -2<br>So B = (2(2-√3), -2)</p><p><strong>Step 4: Calculate area of △AOB</strong></p><p>Using A = (2(2+√3), 2), O = (0,0), B = (2(2-√3), -2)<br>Area = ½|x₁y₂ - x₂y₁| = ½|2(2+√3)(-2) - 2(2-√3)(2)|<br>= ½|-4(2+√3) - 4(2-√3)|<br>= ½|-8-4√3 - 8+4√3|<br>= ½|-16| = 8</p><p><strong>∴ Answer: 8 square units</strong></p>
Correct Answer: 8