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

Jasonbradley

3 min read

·

Jun 17, 2024

27

Share

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.