Problem 31 - 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. $$ x^{2}+y^{2}-4 x-8 y-2=0 $$

Step-by-Step Solution

Answer

The center of the circle is (2, 4) and the radius is 6.

Step 1: Rearrange Terms

First, rearrange the given equation by grouping the x’s and y’s together. Start with: x^{2}-4x + y^{2}-8y - 2 = 0.

Step 2: Complete the Square

To make these groupings into perfect square trinomials, complete the square for both x and y. For the x terms: Take half of -4, the coefficient of x, square it, which is \((-4/2)^2 = 4\). Add and subtract 4 inside the equation: \[(x^2 - 4x + 4) - 4\]For the y terms: Take half of -8, square it, which is \((-8/2)^2 = 16\). Add and subtract 16 inside the equation: \[(y^2 - 8y + 16) - 16\].

Step 3: Simplify the Equation

Now simplify the equation by rewriting it using the perfect square trinomials found in Step 2: ",

Key Concepts

Completing the Square

Completing the square is a technique used to transform a quadratic expression into a perfect square trinomial, which is especially useful for rewriting the equation of a circle. To complete the square, follow these general steps:

• Start with a quadratic term and a linear term, such as \(x^2 - 4x\).
• Focus on the linear term. Take half of its coefficient, then square that number. In this example, take half of \(-4\) which is \(-2\) and square it, resulting in \(4\).
• Add and subtract this square within the expression: \((x^2 - 4x + 4) - 4\). Now, it forms a perfect square trinomial, \((x-2)^2\).

Repeat these steps for the \(y\) terms: \(y^2 - 8y\). Half of \(-8\) is \(-4\), and squaring it produces \(16\). This gives us \((y^2 - 8y + 16) - 16\), simplifying to \((y-4)^2\).Completing the square allows us to rewrite equations in a form that's easy to recognize and solve, such as identifying the properties of a circle.

Graphing Circles

Graphing circles becomes straightforward once we have the circle's equation in the standard form, \((x-h)^2 + (y-k)^2 = r^2\). This form tells us exactly where the circle is located and its size. Here are the steps to graph it:

• The values of \(h\) and \(k\) represent the circle's center. Plot this point on the graph first.
• The circle's radius \(r\) is found by taking the square root of the constant term on the right side of the equation.
• From the center, mark points that are a distance \(r\) away in all four directions (up, down, left, and right).
• Connect these points smoothly to form the circle's outline.

Remember, each point on the curve is equidistant from the center, making it symmetrical and perfectly round. Graphing circles helps visualize their structure and properties.

Center and Radius of a Circle

Identifying the center and radius of a circle from its equation is crucial for graphing and understanding its properties. The equation we use is \((x-h)^2 + (y-k)^2 = r^2\), where:

• \((h, k)\) represents the coordinates of the center of the circle.
• \(r\) is the radius, and it is the square root of the constant on the right side of the equation.

For example, once we complete the square in the provided equation, we derive \((x-2)^2 + (y-4)^2 = 22\). Thus, the center is at \((2, 4)\), and the radius is \(\sqrt{22}\).Understanding these components is essential as it allows you to accurately locate the circle on a graph and understand its size.