Problem 23 - Algebra 2 and Trigonometry

Question

In \(20-27\) : a. Write each equation in center-radius form. b. Find the coordinates of the center. . Find the radius of the circle. $$ x^{2}+y^{2}-6 x+2 y-6=0 $$

Step-by-Step Solution

Answer

Center: (3, -1), Radius: 4.

Step 1: Rearrange the Equation

Start by rearranging the given equation so that the terms involving the same variables are grouped together. The given equation is: \( x^2 + y^2 - 6x + 2y - 6 = 0 \)Rearrange to group the \(x\) and \(y\) terms:\( x^2 - 6x + y^2 + 2y = 6 \)

Step 2: Complete the Square for x-terms

Take the \(x\) terms, \(x^2 - 6x\), and complete the square:1. Take half of the coefficient of \(x\), which is \(-6\), then square it: \((-6/2)^2 = 9\).2. Add and subtract 9 inside the equation: \( x^2 - 6x + 9 \)

Step 3: Complete the Square for y-terms

Take the \(y\) terms, \(y^2 + 2y\), and complete the square:1. Take half of the coefficient of \(y\), which is \(2\), then square it: \((2/2)^2 = 1\).2. Add and subtract 1 inside the equation: \( y^2 + 2y + 1 \)

Step 4: Incorporate the Completed Squares into the Equation

Replace the squared groups in the original rearranged equation:\( (x^2 - 6x + 9) + (y^2 + 2y + 1) = 6 + 9 + 1 \)which simplifies to:\( (x - 3)^2 + (y + 1)^2 = 16 \).

Step 5: Identify Center and Radius

From the equation \( (x - 3)^2 + (y + 1)^2 = 16 \), we extract the center and radius of the circle:1. The center of the circle is \((h, k)\) where \(h = 3\) and \(k = -1\). Thus, the center is \((3, -1)\).2. The radius of the circle is the square root of 16, which is 4.

Key Concepts

Center-Radius Form

A circle's equation can be elegantly represented using the center-radius form. This form is given as \((x-h)^2 + (y-k)^2 = r^2\). Here, \((h,k)\) represents the center of the circle, and \(r\) signifies the radius. This form is particularly useful because it clearly shows the center and radius at a glance, making it easier to understand and visualize the circle on a coordinate plane. When we encounter a circle equation that is not in this form, like \(x^2 + y^2 - 6x + 2y - 6 = 0\), transformations such as completing the square are employed to convert it into the center-radius form.By converting the equation into this form, it becomes much easier to identify the key features of the circle.

Completing the Square

Completing the square is a mathematical technique used to simplify quadratic expressions, and it's extremely handy for transforming circle equations into the center-radius form. This method involves altering a quadratic equation to create a perfect square trinomial, which can then be rewritten as squared binomials.Here's how you do it:

• For the \(x\) terms, we start with \(x^2 - 6x\). Take half of \(-6\) to get \(-3\), then square it to add \(9\). This transforms \(x^2 - 6x\) into \((x-3)^2 - 9\).
• For the \(y\) terms, \(y^2 + 2y\), take half of \(2\) to get \(1\), square it to add \(1\), changing \(y^2 + 2y\) into \((y+1)^2 - 1\).

This method not only simplifies the equation but also sets the stage for easy identification of the circle's properties.

Coordinate Geometry

Coordinate geometry allows us to analyze and describe geometric figures using algebra. When we work with a circle equation on a coordinate plane, we harness the power of algebra to explore geometric properties. This blend of algebra and geometry is crucial for understanding features like distance, points, and shapes in the plane.In our exercise, the circle equation \((x-3)^2 + (y+1)^2 = 16\) showcases how coordinate geometry helps us map out where the circle lies and how it’s structured. We can immediately determine that the circle’s center is located at the point \((3, -1)\), and its radius, derived from the right side of the equation, is \(4\). This fusion of algebraic equations and geometric visualization makes coordinate geometry a powerful tool for problem-solving in mathematics.

Radius Calculation

Calculating the radius is straightforward once the circle’s equation is in the center-radius form \((x-h)^2 + (y-k)^2 = r^2\). Here, \(r^2\) represents the squared radius of the circle. To find the radius, simply solve for \(r\) by taking the square root of \(r^2\).For instance, with our example equation \((x-3)^2 + (y+1)^2 = 16\), we identify that \(r^2 = 16\). The radius \(r\) is obtained by calculating \(\sqrt{16}\), which results in \(r = 4\).Recognizing these values directly from the equation allows us to efficiently and accurately describe and analyze the circle, making radius calculation a fundamental step in understanding the circle's geometric properties.