Circles
Tangent to Circle
Grade 11
Question:
<p>Given three circles of radii \(a\), \(b\), \(c\) \((a < b < c)\) touch each other externally and they have the <em>x</em>-axis as a common tangent. Then:</p>
<p>(A) \(\sqrt{a} = \sqrt{b} + \sqrt{c}\)</p>
<p>(B) \(\sqrt{bc} = \sqrt{ab} + \sqrt{ac}\)</p>
<p>(C) \(\frac{1}{\sqrt{a}} = \frac{1}{\sqrt{b}} + \frac{1}{\sqrt{c}}\)</p>
<p>(D) \(\sqrt{bc} = \sqrt{a}(\sqrt{b} + \sqrt{c})\)</p>
Step-by-Step Solution
Key Concept: When three circles with radii a, b, c are mutually tangent to each other and all tangent to the x-axis, their centers lie on perpendiculars to the x-axis at heights equal to their respective radii. Use the distance formula between centers equals sum of radii for external tangency to establish the relationship.
<p><strong>Step 1:</strong> Set up coordinates. Since all three circles are tangent to the x-axis, place centers at C₁ = (x₁, a), C₂ = (x₂, b), C₃ = (x₃, c) where the y-coordinates equal the radii.</p><p><strong>Step 2:</strong> For circles to be mutually externally tangent: distance between centers = sum of radii.</p><p>Between circles 1 and 2: √[(x₁ - x₂)² + (a - b)²] = a + b</p><p>Squaring: (x₁ - x₂)² + (a - b)² = (a + b)²</p><p>(x₁ - x₂)² = (a + b)² - (a - b)² = 4ab</p><p>So |x₁ - x₂| = 2√(ab)</p><p><strong>Step 3:</strong> Similarly, |x₂ - x₃| = 2√(bc) and |x₁ - x₃| = 2√(ac)</p><p><strong>Step 4:</strong> Arrange circles linearly on x-axis: x₁ < x₂ < x₃</p><p>Then: 2√(ab) + 2√(bc) = 2√(ac)</p><p>Dividing by 2: √(ab) + √(bc) = √(ac)</p><p>Dividing by √(abc): 1/√c + 1/√a = 1/√b</p><p><strong>Therefore:</strong> <strong>1/√a + 1/√c = 1/√b</strong> or equivalently √b = √(ac)/(√a + √c)</p><p>∴ Answer: C</p>
Correct Answer: C