Problem 15 - Algebra and Trigonometry Real Mathematics, Real People

Question

Identify the center and radius of the circle. $$(x-5)^{2}+y^{2}=9$$

Step-by-Step Solution

Answer

The center of the circle is at (5,0) and the radius is 3.

Step 1: Identify the center of the circle

The center of the circle can be found by comparing the given equation with the standard equation. From the equation \((x-5)^{2}+y^{2}=9\), it can be seen that the value of h is 5. Since the equation does not include minus k (since \(y^{2}\) does not have any subtraction or addition), k is equal to 0. So the center of the circle is (5, 0).

Step 2: Identify the radius of the circle

The radius of the circle can be found by looking at the right-hand side of the equation. Our equation's right-hand side is 9 and in standard equation, the right-hand side is \(r^{2}\). This means that \(r^{2}\) is equal to 9. Taking the square root of both sides, we get \(r = \sqrt{9}\), which simplifies to \(r = 3\).

Key Concepts

Circle Center

The center of a circle in an equation plays a pivotal role in determining the circle's position in the coordinate plane. To find the circle center from a circle equation, compare the given equation to the standard form \[(x - h)^2 + (y - k)^2 = r^2\]Here, the center of the circle is represented by the coordinates \((h, k)\).

For example, in the equation \((x-5)^{2}+y^{2}=9\), compare it with the standard form to easily identify \(h\)and \(k\):

• \(h\) = 5,
• \(k\) = 0

Thus, the circle’s center is at the point \((5, 0)\).It's always beneficial for students to highlight the integration and simplification of the standard form in this step, as it clarifies how the center coordinates translate in any circle equation.

Radius of a Circle

Another critical component of a circle is its radius. The radius is crucial because it defines the circle's size by measuring the distance from the center to any point on the circle's border. To understand how to determine a circle's radius from its equation, recognize that in the standard equation formula:\[(x - h)^2 + (y - k)^2 = r^2\]The term \(r^2\)represents the square of the radius.

Looking at the provided example, \((x-5)^{2}+y^{2}=9\), the right-hand side, 9, corresponds to \(r^2\).To find the radius \(r\), take the square root of 9:

• \(r = \sqrt{9} = 3\)

This arithmetic step reveals that the radius is 3. Proper extraction of \(r\) from \(r^2\) through square root operations means that patterns in these equations become more graspable to students, allowing them to see beyond pure memorization.

Standard Form of a Circle Equation

Having a fundamental understanding of the standard form of a circle equation is essential, as it is the template from which students can quickly glean information about a circle's properties. The standard form is expressed as:\[(x - h)^2 + (y - k)^2 = r^2\] Each component of this equation provides insights:

• \(h\)and\(k\):These values identify the circle's center at \((h, k)\).
• \(r^2\):This indicates the square of the circle's radius, allowing you to find \(r\).

Consider breaking equations into these core elements. By doing so, identifying center coordinates and calculating radius become straightforward processes. In any circle equation, these transformations of values provide a consistent, reliable method to analyze and understand circles without confusion. Such simplicity can make mathematics much more accessible to learners.