Problem 23 - Algebra A Combined Function

Question

The graph of each equation is a circle. Find the center and the radius, and then graph the circle. See Examples 5 through 7. $$ x^{2}+y^{2}+6 y=0 $$

Step-by-Step Solution

Answer

The center is \((0, -3)\) and the radius is 3.

Step 1: Identify the equation type

The given equation is a circle equation of the form \(x^2 + y^2 + 6y = 0\). A standard circle equation is given by \((x-h)^2 + (y-k)^2 = r^2\), where \((h, k)\) is the center of the circle and \(r\) is the radius.

Step 2: Rearrange the equation

Rearrange the given equation by isolating \(y\)-related terms. The equation is already in the form where \(x\) is isolated: \(x^2 + (y^2 + 6y) = 0\). Now, we just need to complete the square for the \(y\) terms.

Step 3: Complete the square for the \(y\) terms

Take the \(y\) terms: \(y^2 + 6y\). To complete the square, add and subtract the square of half the coefficient of \(y\). In this case: \(\left(\frac{6}{2}\right)^2 = 9\). So, we write: \(y^2 + 6y + 9 - 9\).

Step 4: Simplify the equation

The equation becomes: \(x^2 + (y^2 + 6y + 9 - 9) = 0\), which simplifies to \(x^2 + (y + 3)^2 - 9 = 0\). Simplify this to \(x^2 + (y+3)^2 = 9\).

Step 5: Identify center and radius

The equation \(x^2 + (y+3)^2 = 3^2\) is now in the standard form \((x-h)^2 + (y-k)^2 = r^2\). Thus, the center \((h, k)\) is \((0, -3)\) and the radius \(r\) is 3.

Step 6: Graph the circle

Graph the circle with the center at \((0, -3)\) and radius 3. This involves plotting the center and using a compass or equivalent tool to mark all points 3 units away from the center, creating a circular shape.

Key Concepts

Completing the Square

Completing the square is a technique used to transform quadratic expressions into a perfect square trinomial form. This method is particularly useful in circle equations to arrange them into a standard form. Here's how it works: When you have a quadratic expression, like the terms in this problem, such as \(y^2 + 6y\), your goal is to make it into something like \((y+k)^2\).
To do this, take the coefficient of the linear term (in this case, 6), halve it to get 3, and then square that result to find the number you need to add and subtract. Thus, you add and subtract 9, giving you \(y^2 + 6y + 9 - 9\). Why do you subtract 9? It's to keep the equation balanced. After completing the square, the expression can be rewritten as \((y+3)^2 - 9\), which then becomes part of the transformed circle equation.

Standard Form of a Circle

The standard form of a circle equation is given by \((x-h)^2 + (y-k)^2 = r^2\).
This format is very convenient because it directly shows you both the center and the radius of the circle without needing additional calculation. In this problem's exercise, we transformed the given equation through completing the square. The original equation \(x^2 + y^2 + 6y = 0\) becomes \(x^2 + (y+3)^2 = 9\).
Here, \((y+3)^2\) matches with \((y-k)^2\) in the standard form equation, showing the transformation and completion of the square. With this standard form, we can easily identify the components needed, like the center and radius of the circle.

Center of a Circle

Finding the center of a circle is straightforward when the equation is in standard form. In the equation \((x-h)^2 + (y-k)^2 = r^2\), the center is given by \((h, k)\). By rewriting our given example as \(x^2 + (y+3)^2 = 9\), it's apparent that our circle's center is \((0, -3)\).
This is because there's no \(x\) term to affect the center in the horizontal direction, so \(h\) is 0, and \(y+3\) translates to \(k = -3\). Knowing the center is crucial. It tells you where to "anchor" your circle on the graph, making it a vital step for any problem involving circles.

Radius of a Circle

The radius of a circle is simply the distance from its center to any point on the circle. In the standard equation form \((x-h)^2 + (y-k)^2 = r^2\), \(r\) is the radius.In our example, once we completed the square and reformed the equation into \(x^2 + (y+3)^2 = 9\), we can see that \(r^2 = 9\).
This means that \(r\), the radius, is the square root of 9, which is 3. The radius gives us the size of the circle; larger values mean a bigger circle. It's also essential for graphing, making sure your circle is the right size on a coordinate plane.