Problem 26 - 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. $$ x^{2}+y^{2}+10 x-5 y-32=0 $$

Step-by-Step Solution

Answer

Center: (-5, 2.5); Radius: \(\sqrt{63.25}\).

Step 1: Rearrange the Equation

Start by organizing the equation into groups of terms: separate the terms involving \(x\) and \(y\).Original equation: \(x^2 + y^2 + 10x - 5y - 32 = 0\).Group \(x\) terms: \((x^2 + 10x)\).Group \(y\) terms: \((y^2 - 5y)\).

Step 2: Complete the Square for x Terms

For the \(x\) terms, \(x^2 + 10x\), complete the square.1. Take the coefficient of \(x\), which is 10.2. Divide by 2 to get 5, then square it to get 25.3. Add and subtract 25 inside the equation to maintain balance: \((x^2 + 10x + 25 - 25)\).

Step 3: Complete the Square for y Terms

For the \(y\) terms, \(y^2 - 5y\), complete the square.1. Take the coefficient of \(y\), which is -5.2. Divide by 2 to get \(-2.5\), then square it to get \(6.25\).3. Add and subtract 6.25 inside the equation to maintain balance: \((y^2 - 5y + 6.25 - 6.25)\).

Step 4: Rewrite the Equation with Completed Squares

Substitute the completed squares back into the rearranged equation.1. From step 2: \((x + 5)^2 - 25)\).2. From step 3: \((y - 2.5)^2 - 6.25)\).3. Substitute back: \((x + 5)^2 - 25 + (y - 2.5)^2 - 6.25 - 32 = 0\).

Step 5: Simplify the Equation

Simplify the equation by combining constants.1. Combine \(-25\), \(-6.25\), and \(-32\) leading to \(-63.25\).2. Rewrite the equation: \((x + 5)^2 + (y - 2.5)^2 = 63.25\).

Step 6: Identify Center and Radius of the Circle

The equation is now in the form \((x - h)^2 + (y - k)^2 = r^2\).1. Compare with \((x + 5)^2 + (y - 2.5)^2 = 63.25\).2. The center \((h, k)\) is \((-5, 2.5)\).3. The radius \(r\) is \(\sqrt{63.25}\).

Key Concepts

Center-Radius Form

The center-radius form of a circle's equation is a standard way to express the characteristics of a circle, making it easier to visualize and understand the geometry involved. The equation takes the form \[ (x - h)^2 + (y - k)^2 = r^2 \] where

• \((h, k)\) represents the center of the circle,
• \(r\) stands for the radius.

This format is particularly useful because it directly gives you the circle's center coordinates and radius. To transition a general equation of a circle into this form, algebraic manipulation techniques such as completing the square are often used. Understanding center-radius form is fundamental because it provides a clear visual representation and simplifies solving geometric problems involving circles.

Completing the Square

Completing the square is an algebraic technique used to simplify equations, particularly when working with quadratic equations. Here’s a simple breakdown of how it works:

• Identify the quadratic and linear terms in the equation: For instance, in \(x^2 + 10x\), the terms are \(x^2\quad \text{and} \quad 10x\).
• Take half of the linear coefficient, square it, and add and subtract this value inside the equation to preserve balance. In our example, half of 10 is 5, and 5 squared is 25, giving us \((x^2 + 10x + 25 - 25)\).

This technique transforms quadratic expressions into perfect square binomials, which are easier to manage and facilitate rewriting the equation in center-radius form.

Circle Center and Radius

Identifying the center and radius of a circle from its equation can be straightforward once it's expressed in center-radius form. Consider the equation \[ (x + 5)^2 + (y - 2.5)^2 = 63.25 \].

• Center: The values \((h, k)\), which correspond to \((-5, 2.5)\), provide the circle's center.
• Radius: The right-hand side of the equation, \(r^2 = 63.25\), allows us to determine \(r\) by taking the square root, yielding \(r = \sqrt{63.25}\).

Understanding how to extract these features from an equation helps in visualizing the circle's position and size, enabling better comprehension of its geometric properties.

Algebraic Manipulation

Algebraic manipulation involves rearranging and simplifying expressions to achieve the desired equation form, essential for solving many math problems. In dealing with circle equations, this often involves grouping terms and adjusting the equation format via techniques like completing the square. These steps help transition from a complex equation to a simpler, more interpretable form, such as the center-radius format. For example,

• Begin by grouping like terms involving \(x\) and \(y\).
• Apply completing the square to convert these into perfect squares.
• Rearrange terms such that constants are isolated on one side, simplifying the equation further.

By mastering these algebraic skills, students can confidently transform and interpret equations to solve geometry and calculus problems effectively.