How to Identify the Conic Section from the Equation x² + 4x + y² + 4y – 4 = 12?

To identify the conic section represented by the equation x² + 4x + y² + 4y – 4 = 12, we first rewrite the equation in a more standard form.

Starting with:

x² + 4x + y² + 4y – 4 = 12

We can rearrange this to:

x² + 4x + y² + 4y = 16

Next, we complete the square for the x and y terms:

• For x² + 4x, we take half of 4 (which is 2), square it (giving us 4), and add/subtract it to complete the square. This gives us:

(x² + 4x + 4) – 4

• For y² + 4y, we do the same thing:

(y² + 4y + 4) – 4

Now, our equation looks like this:

(x + 2)² – 4 + (y + 2)² – 4 = 16

Which simplifies to:

(x + 2)² + (y + 2)² = 24

Now we can clearly see that this is the equation of a circle, given by the standard form (x – h)² + (y – k)² = r². In this case, the center of the circle is at (-2, -2) and the radius r is √24, which simplifies to 2√6.

Thus, the conic section represented by the original equation is a circle.