Circles
Tangent Circles and Descartes Circle Theorem
Grade 11
Question:
<p>Given a line segment AB, A ≡ (0, 0) and B(a, 0). Three circles S₁, S₂, S₃ of radius R are centred at the end points and the midpoint of the line segment AB. A fourth circle S₄ is drawn touching the 3 given circles. If 0 < R < a/4, then sum of all possible distinct values of radius of S₄ is:</p>
<p>(a) \(\frac{a^2}{16R}\)</p>
<p>(b) \(\frac{a^2}{7R}\)</p>
<p>(c) \(\frac{3a^2}{16R}\)</p>
<p>(d) \(\frac{a^2}{4R}\)</p>
Step-by-Step Solution
Key Concept: Tangency conditions between circles lead to distance equations; symmetry of the configuration constrains the possible radii values.
<p><strong>Step 1:</strong> Three circles: S₁ centered at A(0,0) with radius R, S₂ centered at (a/2, 0) with radius R, S₃ centered at B(a, 0) with radius R.</p><p><strong>Step 2:</strong> Circle S₄ with radius r₄ centered at (x, y) must satisfy tangency conditions. Distance from center of S₄ to each of S₁, S₂, S₃ determines whether it's external or internal tangency.</p><p><strong>Step 3:</strong> For 0 < R < a/4, multiple configurations are possible. Using tangency conditions with external contact: |center of S₄ - center of Sᵢ| = r₄ + R (external) or |..| = |r₄ - R| (internal).</p><p><strong>Step 4:</strong> By symmetry and solving the system of distance equations, the sum of all distinct possible radii is \(\frac{a^2}{16R}\).</p><p>∴ Answer is A.</p>
Correct Answer: a