Solving Intersection of Line and Circle in First Quadrant

Problem Setup

• Equation of Line: y = 2x + 1
  • Y-intercept at 1, slope is 2 (rise 2 units, run 1 unit).
• Equation of Circle:
  • Center: (0, 4)
  • Radius: 4
• Aim: Find point of intersection in the first quadrant.

Equation of the Circle

• General form: (x-h)² + (y-k)² = r²
• For given circle:
  • h = 0, k = 4, r = 4
• Circle Equation: x² + (y - 4)² = 16

System of Equations

• Substitute Line Equation in Circle Equation:
  • y = 2x + 1 into x² + (y - 4)² = 16
  • x² + (2x + 1 - 4)² = 16
  • Simplify: x² + (2x - 3)² = 16

Solving the Equation

• Expand and Simplify:
  • x² + (2x-3)(2x-3) = 16
  • x² + 4x² - 12x + 9 = 16
  • 5x² - 12x + 9 - 16 = 0
  • 5x² - 12x - 7 = 0
• Quadratic Equation:
  • Not factorable, use quadratic formula.

Quadratic Formula

• Formula: x = [-b ± sqrt(b² - 4ac)] / 2a
  • a = 5, b = -12, c = -7
• Solve for x in the first quadrant (positive x):
  • x = [12 ± sqrt(144 + 140)] / 10
  • x = [12 ± sqrt(284)] / 10

Finding the X-coordinate

• Calculate: x = (12 + sqrt(284)) / 10
  • Approximation: x ≈ 2.885 (rounded to 3 decimal places)

Finding the Y-coordinate

• Back Substitution: Using y = 2x + 1
  • y = 2(2.885) + 1
  • y ≈ 6.770 (rounded to 3 decimal places)

Conclusion

• Point of Intersection:
  • (x, y) ≈ (2.885, 6.770)
• Verifies with graph: x slightly less than 3, y slightly less than 7.