Circles
Touching Circles
Grade 11

Question:

<p>If \(S_1: x^2 + y^2 = 4\) and \(S_2: x^2 + y^2 - 2ax - 2by + 2 = 0\) touches each other, then 4 times the radius of the circle \(S_2\) is:</p>

Step-by-Step Solution

Key Concept: Two circles touch each other when the distance between their centers equals either the sum (external tangency) or absolute difference (internal tangency) of their radii. Use these conditions to find the relationship between a and b, then calculate the radius of S₂.
<p><strong>Step 1: Identify centers and radii</strong></p><p>S₁: center C₁ = (0, 0), radius r₁ = 2</p><p>S₂: x² + y² - 2ax - 2by + 2 = 0</p><p>Center C₂ = (a, b), radius r₂ = √(a² + b² - 2)</p><p><strong>Step 2: Apply tangency condition</strong></p><p>Distance between centers: d = √(a² + b²)</p><p>For external tangency: √(a² + b²) = 2 + √(a² + b² - 2)</p><p>For internal tangency: √(a² + b²) = |2 - √(a² + b² - 2)|</p><p><strong>Step 3: Solve external tangency case</strong></p><p>Let √(a² + b²) = k, then k = 2 + √(k² - 2)</p><p>k - 2 = √(k² - 2)</p><p>(k - 2)² = k² - 2</p><p>k² - 4k + 4 = k² - 2</p><p>-4k = -6</p><p>k = 3/2 (contradiction since k > 2)</p><p><strong>Step 4: Solve internal tangency case</strong></p><p>k = 2 - √(k² - 2) [taking the valid case]</p><p>√(k² - 2) = 2 - k</p><p>k² - 2 = 4 - 4k + k²</p><p>-2 = 4 - 4k</p><p>4k = 6</p><p>k = 3/2</p><p>But check: √(a² + b²) = 3/2, so a² + b² = 9/4</p><p>r₂ = √(9/4 - 2) = √(1/4) = 1/2</p><p><strong>Step 5: Find 4 times the radius</strong></p><p>4r₂ = 4 × 1/2 = 2</p><p>∴ Answer: <strong>2</strong></p>
Correct Answer: 2

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