Problem 63 - Algebra and Trigonometry Real Mathematics, Real People

Question

Classify the graph of the equation as a circle, a parabola, an ellipse, or a hyperbola. \(x^{2}+y^{2}+2 x-6 y=0\)

Step-by-Step Solution

Answer

The graph of the equation \(x^{2}+y^{2}+2 x-6 y=0\) is a circle with center at (-1,3) and radius \( \sqrt{10} \).

Step 1: Identify the curve type

The given equation is \(x^{2}+y^{2}+2 x-6 y=0\), which can be rewritten as \((x^{2}+2x) + (y^{2} - 6y)=0\). This equation contains both \(x^2\) and \(y^2\) term. Since the coefficients of \(x^2\) and \(y^2\) are both positive and the same, it can be concluded that the graph represents a circle.

Step 2: Convert to standard form

To find the center and radius of the circle, the equation should be written in the standard form \( (x - h)^{2} + (y - k)^{2} = r^{2} \), where (h, k) is the center and r is the radius of the circle. This can be achieved by completing the square. First, re-arrange the equation to \((x^2+2x+1) -1 + (y^2 -6y+9) -9 =0\). Subsequently, the equation can be rewritten as \((x+1)^2 + (y -3)^2 = 1+9 \).

Step 3: Final form and classification

The final equation is \((x+1)^2 + (y-3)^2 =10\). This can be classified as the equation of a circle with center at (-1, 3) and radius \( \sqrt{10} \).

Key Concepts

Standard Form of a Circle

Understanding the standard form of a circle is essential for analyzing and graphing these round wonders of the coordinate plane. It's expressed as \( (x - h)^{2} + (y - k)^{2} = r^{2} \), where \(h, k\) represent the center's coordinates, and \(r\) denotes the radius. This neat package allows us to easily visualize the circle's position in space.

When faced with an equation like \(x^{2}+y^{2}+2x-6y=0\), our goal is to transform it to match the standard form. By meticulously re-arranging terms to complete the square for \(x\) and \(y\), we pave the way to identify the circle's center and radius. In this particular instance, once the squares are completed and constants adjusted, the equation miraculously reveals its true identity as \( (x+1)^{2} + (y-3)^{2} = 10 \), pinpointing the center at (-1,3) with a radius of \(\sqrt{10}\).

Completing the Square

Ever feel like something's missing when you look at a quadratic expression? That's where completing the square comes in—like finding the last puzzle piece to see the whole picture! This technique allows us to transform quadratic equations into a perfect square trinomial, unveiling deeper insights.

For the equation \(x^{2}+y^{2}+2x-6y=0\), we separate the \(x\) and \(y\) terms and hunt for the magic numbers that complete the square. In our example, adding \(1\) to \(x^{2}+2x\) and adding \(9\) to \(y^{2}-6y\) does the trick, forming \(x+1)^{2}\) and \(y-3)^{2}\), respectively.

Remember, though, what we do to one side, we must do to the other to keep the equation balanced. So we adjust by subtracting the same numbers on the other side of the equation, ensuring mathematical harmony and leaving us with the balanced completed-square form.

Graphing Conic Sections

Elbows deep in equations? It's time to graph them and witness conic sections leap off the page and onto the coordinate plane. Whether they're circles, parabolas, ellipses, or hyperbolas, graphing these shapes unravels their mysteries.

With our circle's standard form \( (x+1)^{2} + (y-3)^{2} = 10 \), plotting is a piece of pie (not just a piece, the whole circle!). First, mark the center at (-1,3). Next, wield your compass (or the modern equivalent—a graphing tool) to draw a circle of radius \(\sqrt{10}\), around this central point. Voilà, you've graphed a conic section!

It's like connecting the dots, but with equations and a bit of spatial reasoning. As you practice, you'll find the key features of these shapes—such as vertices, foci, and directrices—fall into place on the graph, turning abstract equations into tangible geometric figures.