(x-3)^2+(y+2)^2=16 2x+2y=10

3 min read Jun 17, 2024
(x-3)^2+(y+2)^2=16 2x+2y=10

Solving a System of Equations: Circle and Line

This article will guide you through solving a system of equations involving a circle and a line. We will analyze the equations (x-3)^2 + (y+2)^2 = 16 and 2x + 2y = 10 to find their points of intersection.

Understanding the Equations

  • (x-3)^2 + (y+2)^2 = 16 represents a circle with:

    • Center: (3, -2)
    • Radius: 4 (square root of 16)
  • 2x + 2y = 10 represents a line. To understand it better, we can rewrite it in slope-intercept form (y = mx + c):

    • Slope (m): -1
    • Y-intercept (c): 5

Solving the System

To find the points of intersection, we need to find the values of x and y that satisfy both equations simultaneously. We can achieve this through the following steps:

  1. Substitution Method:

    • Solve the linear equation (2x + 2y = 10) for one variable (let's solve for x):
      • x = 5 - y
    • Substitute this expression for x into the circle equation:
      • (5 - y - 3)^2 + (y + 2)^2 = 16
      • Simplify and solve the resulting quadratic equation.
  2. Solving the Quadratic Equation:

    • Expanding the equation, we get:
      • (2 - y)^2 + (y + 2)^2 = 16
      • 4 - 4y + y^2 + y^2 + 4y + 4 = 16
      • 2y^2 = 8
      • y^2 = 4
      • y = ±2
  3. Finding the Corresponding x Values:

    • Substitute the values of y back into the equation x = 5 - y:
      • For y = 2, x = 5 - 2 = 3
      • For y = -2, x = 5 - (-2) = 7

Conclusion

Therefore, the points of intersection between the circle (x-3)^2 + (y+2)^2 = 16 and the line 2x + 2y = 10 are:

  • (3, 2)
  • (7, -2)

This means the line intersects the circle at two distinct points.

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