Problem 25 - 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. $$ x^{2}+y^{2}-8 y=0 $$

Step-by-Step Solution

Answer

The center is (0, 4) and the radius is 4.

Step 1: Write the Equation for Rearrangement

Given the equation is \(x^2 + y^2 - 8y = 0\). Our goal is to express it in the form \((x-h)^2 + (y-k)^2 = r^2\).

Step 2: Complete the Square for the y-term

To complete the square, start by focusing on the terms involving \(y\): \(y^2 - 8y\). To form a perfect square trinomial, take half of the coefficient of \(y\), which is \(-8\), divide by 2 to get \(-4\), and then square it, resulting in \(16\). Add and subtract \(16\) within the equation:\[x^2 + (y^2 - 8y + 16) - 16 = 0\].

Step 3: Simplify the Equation

Rewrite the completed square part and move the constant to the other side:\(x^2 + (y-4)^2 = 16\).

Step 4: Compare with Center-Radius Form

Now, compare with the circle equation \((x-h)^2 + (y-k)^2 = r^2\). In this case, \((h, k) = (0, 4)\) and \(r^2 = 16\).

Step 5: Determine the Circle's Center and Radius

The center of the circle is \((h, k) = (0, 4)\). The radius \(r\) is the square root of \(16\), which is \(4\).

Key Concepts

Center-Radius Form

The center-radius form of a circle's equation looks like this: \((x-h)^2 + (y-k)^2 = r^2\). Here, \((h, k)\) represents the center's coordinates, and \(r\) is the radius of the circle. This form is essential as it clearly gives us both the center coordinates and radius just by looking at the equation.
To convert an equation to this form, we may need to rearrange and manipulate given terms, often using a method like completing the square. Having equations in this form simplifies understanding and analyzing circles in geometry, as you can easily identify how the circle is positioned on a graph.

Completing the Square

Completing the square is a mathematical technique that transforms a quadratic expression into a perfect square trinomial. This method is crucial in rewriting circle equations into the center-radius form.

• First, focus on the quadratic terms that need transforming—often the \(x\) or \(y\) terms, as seen in \(y^2 - 8y\).
• Next, take the coefficient of the linear term (in our case, \(-8\)), divide it by 2 to get \(-4\), and square it, resulting in \(16\).
• Insert this squared number into the equation, adding and subtracting within the same step, to maintain equality.
• This process results in a perfect square trinomial, \((y-4)^2\), making our equation easier to manage.

This approach helps in neatly restructuring the equation to reveal important features of a circle, like its center and radius.

Coordinates of the Center

The coordinates \((h, k)\) in the center-radius form \((x-h)^2 + (y-k)^2 = r^2\) stand for the center of the circle. Accurately determining these allows us to precisely locate the circle on a plane or graph.
In our example, from the equation \((x-0)^2 + (y-4)^2 = 16\), the center is clearly \((0, 4)\). This extraction comes directly from comparing the modified equation with the standard form.
Understanding and finding the center coordinates are foundational for graphing circles and for solving more complex geometric problems involving circles.

Radius of a Circle

In the center-radius form, the radius \(r\) is pivotal as it dictates the size of the circle. It's the distance from the center \((h, k)\) to any point on the circle's boundary.
Finding the radius involves taking the square root of the number on the right-hand side of the equation \(r^2\), which we've rearranged into the center-radius form.
In our study of the example equation \(x^2 + (y-4)^2 = 16\), \(r^2\) equals 16. Thus, \(r\) is \(\sqrt{16} = 4\).
Knowing \(r\) not only specifies the circle's size but is also instrumental in many applications, both in pure mathematics and in practical geometric computations.