How do I find the center and radius of the circle defined by the equation x² + 4x + y² + 8y + 4 = 0?

Finding the Center and Radius of the Circle

To find the center and radius of the circle from the given equation, we need to rewrite the equation in the standard form of a circle, which is:

(x – h)² + (y – k)² = r²

where (h, k) is the center of the circle and r is the radius.

Step 1: Rewrite the equation

The original equation is:

x² + 4x + y² + 8y + 4 = 0

First, we move the constant term to the right side of the equation:

x² + 4x + y² + 8y = -4

Step 2: Complete the square

Next, we will complete the square for the x and y terms.

1. For the x terms: x² + 4x

  • Take half of the coefficient of x (which is 4), square it, and add it: (4/2)² = 4.
  • Add and subtract this value inside the equation.

2. For the y terms: y² + 8y

  • Take half of the coefficient of y (which is 8), square it, and add it: (8/2)² = 16.
  • Add and subtract this value inside the equation as well.

So, we rewrite the equation as:

(x² + 4x + 4) + (y² + 8y + 16) = -4 + 4 + 16

This simplifies to:

(x + 2)² + (y + 4)² = 16

Step 3: Identify the center and radius

Now we can identify the center and the radius from our completed equation:

  • The standard form is (x – h)² + (y – k)² = r².
  • Comparing, we see: h = -2, k = -4, and r² = 16.
  • Thus, the center (h, k) is (-2, -4) and the radius r is √16 = 4.

Final Result

The center of the circle is (-2, -4) and the radius is 4.

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