Problem 24 - Algebra 2

Question

Find the center and radius of the circle with the given equation. Then graph the circle. $$ (x-3)^{2}+y^{2}=16 $$

Step-by-Step Solution

Answer

The center of the circle is (3, 0) and the radius is 4.

Step 1: Identify the standard form of the circle equation

The general equation of a circle in standard form is \((x-h)^2 + (y-k)^2 = r^2\), where \((h, k)\) is the center of the circle and \(r\) is the radius. Compare this with the given equation \((x-3)^2 + y^2 = 16\).

Step 2: Determine the center of the circle

In the equation \((x-3)^2 + y^2 = 16\), compare with the standard form \((x-h)^2 + (y-k)^2 = r^2\). You can see that \(h = 3\) and since there is no \(k\) value explicitly written, \(k = 0\). Thus, the center of the circle is \((3, 0)\).

Step 3: Calculate the radius of the circle

From the equation \((x-3)^2 + y^2 = 16\), identify that \(r^2 = 16\). Solving for \(r\), the radius, we find \(r = \sqrt{16} = 4\).

Step 4: Graph the circle

To graph the circle, plot the center point \((3, 0)\) on the coordinate plane. Then, draw a circle with a radius of 4 around this center. Ensure that the circle crosses 4 units from the center point in all directions: left, right, up, and down.

Key Concepts

center of a circle

The center of a circle is a key concept in understanding its equation. Imagine a point that sits right in the middle of a perfectly round circle. This point is called the 'center.' It determines the exact point from which every edge of the circle is equidistant.

In mathematical terms, the center of a circle is represented by the coordinates \(h, k\). In the equation of the circle, \((x-h)^2 + (y-k)^2 = r^2\), \(h\) and \(k\) are the values that make up these coordinates, where:

• \(h\) represents the x-coordinate of the center.
• \(k\) represents the y-coordinate of the center.

Each coordinate value shows how far off the circle's center is from the origin point (0, 0) on a graph.

For instance, in the equation \((x-3)^2 + y^2 = 16\), the coordinates of the center are determined to be \(3, 0\). This means the circle's center is at 3 units along the x-axis and exactly on the y-axis.

radius of a circle

The radius of a circle is the constant distance from its center to any point on its edge. Think of this distance as a string that you can stretch exactly from the center to the boundary of the circle.

In the circle equation, \((x-h)^2 + (y-k)^2 = r^2\), the radius is represented by \(r\). However, in the formula, it appears as \(r^2\), meaning that you need to take the square root to find the actual length of the radius:

• Identify \(r^2\) from the equation.
• Calculate \(r\) by taking the square root of \(r^2\).

In our example, since the equation is \((x-3)^2 + y^2 = 16\), we learn that \(r^2 = 16\). By calculating \(r\), or the square root of 16, we determine that the radius is 4. Thus, every point on our circle's boundary is precisely 4 units away from its center at \(3, 0\).

standard form of a circle equation

The standard form of a circle equation is a specific way to express a circle mathematically. Using this form allows one to quickly identify key properties, like the circle's center and radius.

The standard form equation is given by \((x-h)^2 + (y-k)^2 = r^2\). It associates any x-value and y-value that satisfy this equation with points on the circle.

• \(x\) and \(y\) are variables representing any point on the circle.
• \(h\) and \(k\) denote the center coordinates of the circle, as seen previously.
• \(r\) indicates the circle's radius.

One crucial benefit of using the standard form is its ease of comparison to deduce both the center and radius at a glance. For example, comparing the standard form to our equation \((x-3)^2 + y^2 = 16\) quickly reveals that the center \(h, k\) is \(3, 0\), and \(r^2 = 16\). The process facilitates solving, plotting, and understanding the behavior of the circle in different scenarios.