Notes on Circle Chord Problem by Presh Talwalkar

Introduction

• Speaker: Presh Talwalkar
• Problem description:
  • Large circle with chord PQ (length = 6)
  • Two small circles:
    • One above chord (radius = a)
    • One below chord (radius = b)
  • Both small circles tangent to chord and large circle, touching each other
• Goal: Find the shaded area between the two small circles and the large circle.

Solution Methods

Method 1: Right Triangle Approach

1. Definitions:

   • Large circle radius = r
   • Diameter of top small circle = 2a
   • Diameter of bottom small circle = 2b
   • Relationship:
     • 2r = 2a + 2b
     • Therefore, r = a + b

2. Area Calculation:

   • Shaded area = Area of large circle - Area of small circles
   • Formula:
     • Shaded area = πr² - πa² - πb²
   • Substitute r = a + b:
     • Shaded area = π(a + b)² - πa² - πb²
   • Expand binomial and simplify:
     • Resulting shaded area = 2abπ

3. Distance Calculation:

   • Distance from center of large circle to chord PQ:
     • r - 2a = (a + b) - 2a = b - a
   • Right triangle formed:
     • One leg = b - a
     • Hypotenuse = a + b
     • Other leg = 3 (half of PQ)

4. Right Triangle Equation:

   • From Pythagorean theorem:
     • (a + b)² = (b - a)² + 3²
   • Simplifying gives:
     • 4ab = 9
   • Therefore, shaded area = 2abπ = (9/2)π

Method 2: Chord-Chord Power Theorem

1. Setup:

   • Diameter of large circle endpoints: s and t, intersects chord PQ at m
   • Apply chord-chord power theorem:
     • sm * mt = pm * mq

2. Substitutions:

   • SM = 2A
   • MT = 2B
   • PM = MQ = 3
   • This leads to:
     • 2A * 2B = 3 * 3
     • Resulting in 4AB = 9

3. Final Calculation:

   • Shaded area = 2ABπ = (9/2)π

Conclusion

• Both methods lead to the same result: Shaded area = 9π/2
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