<p>The radius of a circle, having minimum area, which touches the curve \(y = 4 - x^2\) and the lines \(y = |x|\) is</p>
<p>\(2(\sqrt{2} - 1)\)</p>
<p>\(4(\sqrt{2} - 1)\)</p>
<p>\(4(\sqrt{2} + 1)\)</p>
<p>\(2(\sqrt{2} + 1)\)</p>
Step-by-Step Solution
Key Concept: A circle of minimum area touching the parabola y = 4 - x² and both lines y = |x| must be centered on the y-axis by symmetry, and the optimal configuration occurs when the circle is tangent to the parabola at its vertex and tangent to both lines simultaneously.
<p><strong>Step 1: Set up the geometry using symmetry</strong></p><p>By symmetry of the parabola y = 4 - x² and lines y = |x|, the circle of minimum area must be centered on the y-axis at some point (0, h) with radius r.</p><p><strong>Step 2: Apply tangency condition to y = |x|</strong></p><p>For the circle centered at (0, h) with radius r to be tangent to the line y = x (valid for x > 0), the distance from (0, h) to the line x - y = 0 equals r:</p><p>$$r = \frac{|0 - h|}{\sqrt{2}} = \frac{h}{\sqrt{2}}$$</p><p>Therefore: $h = r\sqrt{2}$</p><p><strong>Step 3: Apply tangency condition to y = 4 - x²</strong></p><p>The circle must touch the parabola. By symmetry, contact occurs at the vertex (0, 4). The distance from center (0, h) to point (0, 4) equals r:</p><p>$$|4 - h| = r$$</p><p>Since the circle is below the parabola's vertex: $4 - h = r$, so $h = 4 - r$</p><p><strong>Step 4: Solve for r</strong></p><p>From Steps 2 and 3:</p><p>$$r\sqrt{2} = 4 - r$$</p><p>$$r\sqrt{2} + r = 4$$</p><p>$$r(\sqrt{2} + 1) = 4$$</p><p>$$r = \frac{4}{\sqrt{2} + 1}$$</p><p><strong>Step 5: Rationalize the denominator</strong></p><p>$$r = \frac{4}{\sqrt{2} + 1} \cdot \frac{\sqrt{2} - 1}{\sqrt{2} - 1} = \frac{4(\sqrt{2} - 1)}{2 - 1} = 4(\sqrt{2} - 1)$$</p><p><strong>Verification:</strong> With r = 4(√2 - 1):</p><p>- h = r√2 = 4(√2 - 1)√2 = 4(2 - √2)</p><p>- Distance to (0,4): |4 - 4(2 - √2)| = |4√2 - 4| = 4(√2 - 1) = r ✓</p><p><strong>∴ Answer: B</strong></p>
Correct Answer: B