Problem 59 - Precalculus Essentials

Question

Complete the square and write the equation in standard form. Then give the center and radius of each circle and graph the equation. $$x^{2}-2 x+y^{2}-15=0$$

Step-by-Step Solution

Answer

The given equation is equivalent to \( (x-1)^{2} + y^{2} = 16 \) in standard form. The center of the circle is (1,0) and the radius is 4.

Step 1: Convert to standard form

To rewrite in the standard form we need to group the x's and y's together. The given equation can be rewritten as: \( x^{2}-2x + y^{2} = 15 \), now we just have to complete the square for the x's. In order to do that, we have to find \((b/2)^{2}\), where b is the coefficient of x, which is -2. This gives us \(((-2)/2)^{2} = 1\), which we add to both sides to complete the square: \( (x^{2}-2x+1) + y^{2} = 15 + 1 \). The equation simplifies to \( (x-1)^{2} + y^{2} = 16 \).

Step 2: Identify the center and the radius

From the standard form of the circle equation, we can see that the circle's center is at (1, 0) and the radius of the circle is the square root of 16 which is 4. So, the center is (1,0) and the radius is 4.

Key Concepts

Completing the Square

Completing the square is a technique used to simplify and rewrite quadratic expressions. It rearranges the expression into a perfect square trinomial, which makes it easier to solve or graph the equation. In our exercise, we had to complete the square for the expression involving the variable \( x \). Here’s how it works step by step:

• First, ensure all related terms are grouped together: \( x^2 - 2x + y^2 = 15 \).
• Identify the coefficient of \( x \) which is \( -2 \). Use this to find \( (b/2)^2 \): Here, \( b = -2 \), so \( ((-2)/2)^2 = 1 \).
• Add \( 1 \) to both sides to balance the equation: \( (x^2 - 2x + 1) + y^2 = 16 \).
• Now rewrite \( x^2 - 2x + 1 \) as \( (x-1)^2 \). This represents a perfect square trinomial.

By completing the square, we transition from a general quadratic form to one that easily displays the characteristics we need for further analysis.

Standard Form of a Circle

The standard form of a circle’s equation is crucial for identifying its center and radius quickly. The general standard form is \( (x-h)^2 + (y-k)^2 = r^2 \), where \((h, k)\) represents the center of the circle and \( r \) is the radius.After completing the square in the exercise, we obtained the equation \( (x-1)^2 + y^2 = 16 \). Comparing this with the standard form allows us to glean important information:

• \( h = 1 \) and \( k = 0 \), designating \( (1, 0) \) as the center of the circle.
• \( r^2 = 16 \), giving us a radius \( r = 4 \) by taking the square root of 16.

The equation’s standard form makes these components visually clear, simplifying both analysis and geometric representation of the circle.

Center and Radius of a Circle

Identifying the center and radius of a circle from its equation is direct once it is in standard form. The variables \( h \) and \( k \) in the equation \( (x-h)^2 + (y-k)^2 = r^2 \) correspond to the \( x \) and \( y \) coordinates of the circle’s center, \( (h, k) \). The term \( r^2 \) equates to the square of the radius. Therefore, to find the radius \( r \), we simply take the square root of \( r^2 \). These elements not only define the circle’s size and position but are integral for sketching its graph.In our exercise, the equation \( (x-1)^2 + y^2 = 16 \) informs us that:

• The center, \( (h, k) \), is \( (1, 0) \).
• The radius \( r \) is \( 4 \), derived from \( r^2 = 16 \) where \( r = \sqrt{16} = 4 \).

This understanding facilitates the process of graphing the circle accurately, as the center's coordinates and radius are precisely defined.