Circles
Tangent to Circle
Grade 11
Question:
<p>From the following figure, which depicts the given situation, the circle touches the line \(y = -x\). Also, from the graph, the radius is obtained as \(4 - k\). The circle passes through \((0, 4)\) and its centre is at \((0, k)\). The radius of the circle is:</p>
<p>\(4(\sqrt{2} + 1)\)</p>
<p>\(4(\sqrt{2} - 1)\)</p>
<p>\(2(\sqrt{2} - 1)\)</p>
<p>\(2(\sqrt{2} + 1)\)</p>
Step-by-Step Solution
Key Concept: Use the distance formula from center to tangent line equals radius, combined with the condition that the circle passes through (0,4). The center at (0,k) and point (0,4) give one constraint, while tangency to y=-x gives another.
<p><strong>Step 1:</strong> Since the circle passes through (0,4) and has center at (0,k), the radius is:</p><p>r = √[(0-0)² + (4-k)²] = |4-k|</p><p><strong>Step 2:</strong> Since the circle touches the line y = -x (or x+y=0), the perpendicular distance from center (0,k) to this line equals the radius:</p><p>Distance = |0+k|/√(1²+1²) = |k|/√2</p><p><strong>Step 3:</strong> Equating both expressions for radius:</p><p>|4-k| = |k|/√2</p><p><strong>Step 4:</strong> Assuming k > 0 and k < 4 (from geometry), we have 4-k = k/√2</p><p>Multiply by √2: (4-k)√2 = k</p><p>4√2 - k√2 = k</p><p>4√2 = k(1 + √2)</p><p>k = 4√2/(1+√2) = 4√2(1-√2)/[(1+√2)(1-√2)] = 4√2(1-√2)/(1-2) = 4√2(√2-1) = 4(2-√2) = 8-4√2</p><p><strong>Step 5:</strong> Therefore, radius = 4-k = 4-(8-4√2) = 4√2-4 = 4(√2-1)</p><p>∴ Answer: B</p>
Correct Answer: B