Problem 35 - Algebra A Combined Function

Question

Hint: For Exercises 33 through 38 , first divide the equation through by the coefficient of \(x^{2}\) (or \(\left.y^{2}\right)\). $$ 6(x-4)^{2}+6(y-1)^{2}=24 $$

Step-by-Step Solution

Answer

The center is (4, 1) and the radius is 2.

Step 1: Divide by the Coefficient

First, divide the entire equation by the coefficient of the term \(6\), which is present with \((x-4)^2\) and \((y-1)^2\), to simplify the equation.\[ \frac{6(x-4)^2}{6} + \frac{6(y-1)^2}{6} = \frac{24}{6} \] This simplifies to \( (x-4)^2 + (y-1)^2 = 4 \).

Step 2: Identify the Standard Form

Recognize that we now have an equation of the form \( (x-h)^2 + (y-k)^2 = r^2 \), which is the standard form of a circle. Here, \(h = 4\), \(k = 1\), and \(r^2 = 4\).

Step 3: Determine the Center and Radius

From the standard form \( (x-h)^2 + (y-k)^2 = r^2 \), identify the center \((h, k)\) and the radius \(r\). Thus, the center is \((4, 1)\) and radius \(r = \sqrt{4} = 2\).

Key Concepts

Standard Form

When dealing with the equations of circles, the standard form is an essential concept. This form expresses the equation of a circle in such a way that makes it easy to identify important features like the center and radius.
The standard form of a circle is given by \[ (x-h)^2 + (y-k)^2 = r^2 \]where

• h is the x-coordinate of the center of the circle,
• k is the y-coordinate of the center, and
• r is the radius of the circle.

This form enables you to see, at a glance, the circle's center located at \((h, k)\), and its radius, which is simply the square root of the term on the right side of the equation.

Radius

The radius of a circle is a measure of the distance from its center to any point on its circumference. In the standard circle equation \((x-h)^2 + (y-k)^2 = r^2\), the term \(r^2\) represents the square of the radius.
To find the radius, you must take the square root of \(r^2\). This is because \(r^2\) is the radius value squared.
For example, if in the equation \((x-4)^2 + (y-1)^2 = 4\), notice that \(r^2 = 4\). Therefore, the radius \(r\) is \(\sqrt{4} = 2\). The radius is a constant length throughout the circle and is crucial in understanding the geometry and size of the circle.

Center of Circle

The center of a circle is a point from which every point on the circle is equidistant. In a circle's standard equation, \((x-h)^2 + (y-k)^2 = r^2\), the center is easily found at \((h, k)\). This makes identifying and plotting the center straightforward.
For the given equation \((x-4)^2 + (y-1)^2 = 4\), \(h\) and \(k\) can be directly read off from the equation, giving the center as \((4, 1)\). This concept provides the anchor around which the circle is drawn.
The center plays a pivotal role not only in drawing the circle but also in various applications involving shifts or movements of the circle in a coordinate plane. Understanding where the center is allows one to easily manipulate the circle's position in mathematical problems.