Equation and Conic Sections

Equations involving $x$ and $y$ can represent different shapes in a 2D plane. These shapes are often referred to as conic sections because they can be formed by intersecting a cone with a plane. Common conic sections include parabolas, circles, ellipses, and hyperbolas. Sometimes, an equation might represent a degenerate conic section, such as a pair of straight lines, a single point, or an empty set.

Identifying Conic Section Type

The general equation for a second-degree curve in two variables $x$ and $y$ is given by:

$\text{Ax}^2 + \text{Bxy} + \text{Cy}^2 + \text{Dx} + \text{Ey} + \text{F} = 0$

The type of conic section represented by this equation can often be determined by the coefficients A, B, and C.

• If $\text{B}^2 - 4\text{AC} = 0$, the equation represents a parabola (or degenerate parabola).
• If $\text{B}^2 - 4\text{AC} < 0$, the equation represents an ellipse or a circle (or a degenerate ellipse like a point). A special case is when A = C and B = 0, which specifically indicates a circle.
• If $\text{B}^2 - 4\text{AC} > 0$, the equation represents a hyperbola (or degenerate hyperbola like a pair of intersecting lines).

Let's look at the given equation: $x^2 + y^2 + 2x = 0$.

Comparing this to the general form, we have:

• A = 1 (coefficient of $x^2$)
• B = 0 (coefficient of $xy$)
• C = 1 (coefficient of $y^2$)
• D = 2 (coefficient of $x$)
• E = 0 (coefficient of $y$)
• F = 0 (constant term)

First, let's calculate $\text{B}^2 - 4\text{AC}$:

$\text{B}^2 - 4\text{AC} = (0)^2 - 4(1)(1) = 0 - 4 = -4$

Since $\text{B}^2 - 4\text{AC} = -4 < 0$, the equation represents either an ellipse or a circle (or a degenerate ellipse). Because A = 1 and C = 1 and B = 0, this is the specific condition for a circle.

Circle Standard Form

The standard form of the equation of a circle with center $(\text{h}, \text{k})$ and radius $r$ is:

$(x - \text{h})^2 + (y - \text{k})^2 = r^2$

Let's rearrange the given equation $x^2 + y^2 + 2x = 0$ to match this standard form by completing the square for the $x$ terms.

Group the $x$ terms together: $(x^2 + 2x) + y^2 = 0$

To complete the square for $x^2 + 2x$, take half of the coefficient of $x$ (which is 2), square it $((2/2)^2 = 1^2 = 1)$, and add it inside the parenthesis. To keep the equation balanced, add it to both sides.

$(x^2 + 2x + 1) + y^2 = 0 + 1$

Now, the expression in the parenthesis is a perfect square: $(x + 1)^2$. The $y^2$ term can be written as $(y - 0)^2$.

$(x + 1)^2 + (y - 0)^2 = 1$

This equation is now in the standard form of a circle:

$(x - (-1))^2 + (y - 0)^2 = 1^2$

Comparing this to $(x - \text{h})^2 + (y - \text{k})^2 = r^2$, we can see that:

• $\text{h} = -1$
• $\text{k} = 0$
• $r^2 = 1$, so $r = \sqrt{1} = 1$

The equation represents a circle with its center at $(-1, 0)$ and a radius of 1.

Conclusion

Based on the analysis of the equation $x^2 + y^2 + 2x = 0$, both by examining the coefficients and by rearranging it into standard form, it is confirmed that the equation represents a circle.