Problem 73 - Intermediate Algebra

Question

Sketch the graph of each equation. If the graph is a parabola, find its vertex. If the graph is a circle, find its center and radius. $$x^{2}-10 y+y^{2}+4=0$$

Step-by-Step Solution

Answer

The graph is a circle with center (0, 5) and radius \(\sqrt{21}\).

Step 1: Identify the Equation Type

The given equation is \(x^{2} - 10y + y^{2} + 4 = 0\). We need to rewrite it in a recognizable form to identify whether it is a parabola or a circle.

Step 2: Rearrange the Terms

Rearrange the equation terms to group the quadratic terms together: \(x^{2} + y^{2} - 10y = -4\).

Step 3: Complete the Square for y

To complete the square for the \(y\) terms, start with \(y^{2} - 10y\). Take half of -10, which is -5, square it to get 25, and add and subtract it inside the equation: \(x^{2} + (y^{2} - 10y + 25) = 25 - 4\).

Step 4: Simplify the Equation

The equation becomes \(x^{2} + (y - 5)^{2} = 21\). This matches the standard form of a circle \((x - h)^{2} + (y - k)^{2} = r^{2}\).

Step 5: Identify the Circle's Center and Radius

The transformed equation \((x - 0)^{2} + (y - 5)^{2} = 21\) indicates the center is at \((0, 5)\) and the radius is \(\sqrt{21}\).

Key Concepts

Completing the Square

Completing the square is a mathematical method used to transform quadratic expressions into a more manageable form. This technique is particularly useful for rewriting equations of circles and parabolas in their standard forms. To complete the square for a quadratic expression like \(y^2 - 10y\), follow these steps:

• Take the coefficient of the linear term, which in this case is -10.
• Divide it by 2 to get -5.
• Square the result to get 25.

Then, add and subtract this square (25) within the expression to keep the equation balanced. This step effectively transforms \(y^2 - 10y\) into \((y - 5)^2 - 25\), simplifying it into a perfect square form \((y - 5)^2\). Completing the square is essential for converting a general quadratic equation into its standard form, making it easier to identify features like the center and radius of a circle.

Center of a Circle

The center of a circle is a crucial element in understanding its geometry. In the standard circle equation form \((x-h)^2 + (y-k)^2 = r^2\), the coordinates \((h, k)\) represent the center of the circle. When you rearrange a general equation to this form, you can easily identify the circle's center.
In the transformed equation we obtained, \((x - 0)^2 + (y - 5)^2 = 21\), the center is easily identifiable as \((0, 5)\). Knowing the center allows you to locate the circle precisely on the Cartesian plane. Always ensure to adjust any rearrangements of terms carefully to accurately find the center coordinates.

Radius of a Circle

Determining the radius of a circle follows directly from its equation. The standard format for a circle’s equation is \((x - h)^2 + (y - k)^2 = r^2\), where \(r^2\) is the square of the radius.
After completing the square and rewriting the equation as \((x - 0)^2 + (y - 5)^2 = 21\), you find that \(r^2 = 21\). To find the radius, simply take the square root of 21.

• The radius \(r = \sqrt{21}\).

Understanding how to express the radius helps determine the size of the circle, providing insight into its spatial characteristics. Always check your arithmetic to ensure accuracy when extracting the radius from the equation.

Quadratic Terms

Quadratic terms in an equation are characterized by variables raised to the power of two, such as \(x^2\) and \(y^2\). These terms are the hallmark of quadratic equations. In circle equations, they are essential because they define the circle's shape and size.
In our given equation, the quadratic terms are \(x^2\) and \(y^2\). By grouping these terms correctly, you can apply techniques like completing the square to manipulate the equation into its standard form.

• Recognizing the separate quadratic contributions is crucial for accurately adjusting the equation.

These terms must be balanced accurately during any transformations, ensuring that you reach the standard circle form without altering the equation's overall context.