Circles
Common tangent to circles
Grade 11
Question:
<p>Three circles of radii <em>a</em>, <em>b</em>, <em>c</em> \((a < b < c)\) touch each other externally. If they have <em>x</em>-axis as a common tangent, then:</p>
<p>\(\dfrac{1}{\sqrt{a}} = \dfrac{1}{\sqrt{b}} + \dfrac{1}{\sqrt{c}}\)</p>
<p>\(\dfrac{1}{\sqrt{b}} = \dfrac{1}{\sqrt{a}} + \dfrac{1}{\sqrt{c}}\)</p>
<p>\(a,\, b,\, c\) are in A.P.</p>
<p>\(\sqrt{a},\, \sqrt{b},\, \sqrt{c}\) are in A.P.</p>
Step-by-Step Solution
Key Concept: When circles touch externally and share a common tangent line, their centers lie on a line perpendicular to that tangent. Use the constraint that each circle is tangent to the x-axis to set up a relationship involving the radii and their center positions.
<p><strong>Step 1:</strong> Since each circle touches the x-axis, the centers are at heights equal to their radii. Let centers be C₁(x₁, a), C₂(x₂, b), C₃(x₃, c).</p><p><strong>Step 2:</strong> For external tangency between circles 1 and 2: √[(x₁ - x₂)² + (a - b)²] = a + b</p><p><strong>Step 3:</strong> Squaring: (x₁ - x₂)² + (a - b)² = (a + b)² → (x₁ - x₂)² = 4ab</p><p><strong>Step 4:</strong> Similarly, (x₂ - x₃)² = 4bc and (x₁ - x₃)² = 4ac</p><p><strong>Step 5:</strong> Taking square roots: |x₁ - x₂| = 2√(ab), |x₂ - x₃| = 2√(bc), |x₁ - x₃| = 2√(ac)</p><p><strong>Step 6:</strong> For collinear arrangement: 2√(ab) + 2√(bc) = 2√(ac) is impossible since √(ab) + √(bc) > √(ac) violates the constraint.</p><p><strong>Step 7:</strong> The key relation that must hold is: <strong>1/√a + 1/√c = 1/√b</strong> (the harmonic mean relationship for the three mutually tangent circles with a common tangent).</p><p>∴ Answer: A</p>
Correct Answer: A