Problem 67 - Intermediate Algebra

Question

Write each equation in standard form, if it is not already so, and graph it. If the graph is a circle, give the coordinates of its center and its radius. If the graph is a parabola, give the coordinates of its vertex. $$ (x-1)^{2}+(y-3)^{2}=15 $$

Step-by-Step Solution

Answer

The equation represents a circle with center (1, 3) and radius \(\sqrt{15}\).

Step 1: Identify the Given Equation

The given equation is \((x-1)^{2}+(y-3)^{2}=15\). This is a standard form equation of a circle.

Step 2: Express in Standard Circle Form

The standard form of a circle is \((x-h)^{2} + (y-k)^{2} = r^{2}\), where \((h, k)\) is the center, and \(r\) is the radius. From \((x-1)^{2}+(y-3)^{2}=15\), we identify that \(h=1\), \(k=3\), and \(r^{2}=15\).

Step 3: Identify the Center and Radius

The center of the circle is obtained from \((h, k)\), which is \((1, 3)\). The radius \(r\) can be found by taking the square root of 15, so \(r=\sqrt{15}\).

Step 4: Write the Circle Equation in Standard Form

The equation \((x-1)^{2}+(y-3)^{2}=15\) is already in "standard form" for a circle. Thus, no transformation is necessary to write it in standard form.

Step 5: Graph the Circle

Plot the center of the circle on a coordinate plane at the point \((1, 3)\). Using the radius \(\sqrt{15}\), draw a circle around the center. This graph represents all points \((x, y)\) that satisfy the equation.

Key Concepts

Understanding Circle Equation Standard Form

The circle equation standard form is fundamental to identifying and graphing circles in algebra and geometry. The equation is expressed as

• \((x-h)^{2} + (y-k)^{2} = r^{2}\)

Here,

• \((h, k)\) refers to the center of the circle,
• \(r\) is the radius of the circle.

This format allows us to easily pinpoint both the circle's center and its radius just by inspecting the equation. It shows that a circle is made up of all points that are equidistant (the distance being the radius) from the given center. Whenever you're given an equation in this format, you can quickly determine these key features of the circle without additional calculations, simplifying the process of analysis and graphing.

Identifying the Center of a Circle

When analyzing a circle’s equation in standard form, identifying the center is straightforward. The center, denoted

• \((h, k)\) in the equation \((x-h)^{2} + (y-k)^{2} = r^{2}\),

is the point around which the circle is perfectly symmetrical.
In our example equation, \((x-1)^{2}+(y-3)^{2}=15\), observing the expressions \( (x-1)^2 \) and \( (y-3)^2 \) directly indicates that the center of the circle is at

• \((1, 3)\).

This point acts as the pivot for drawing the circle, and all other points on the circle are uniformly spaced from this location.

Finding the Radius of a Circle

The radius of a circle is the distance from the center to any point on the circle. In the circle equation standard form

• \((x-h)^{2} + (y-k)^{2} = r^{2}\),

\(r^{2}\) is the squared radius.
To find the actual radius, we need to calculate the square root of the constant term on the right side of the equation.
In the equation \((x-1)^{2}+(y-3)^{2}=15\), we have

• \(r^{2} = 15\),

so the radius \(r\) is

• \(\sqrt{15}\).

Thus, the radius is an irrational number approximately equal to 3.87, which can be used when graphing or applying the circle in real-world contexts.

Graphing a Circle

Graphing a circle involves plotting points on a coordinate plane based on the center and radius obtained from the circle's equation. To graph

• \((x-1)^{2}+(y-3)^{2}=15\),

follow these steps:

• Start by plotting the center of the circle at point \((1, 3)\) on the graph.
• From this center, use a compass or a measuring tool to spread \(\sqrt{15}\) units (approximately 3.87 units) in all directions. This step gives the precise circumference of the circle.
• Draw a round shape connecting all these measured points around the center, ensuring the shape is even and symmetrical.

When completed, the graph will show a perfect circle, showcasing the locations of all the points that meet the equation's condition. This visual representation helps in better understanding circle properties and allows for practical application in various problems.