Problem 59 - Calculus for Biology and Medicine

Question

Find the center and the radius of the circle given by the equation $$ (x-2)^{2}+y^{2}=16 $$

Step-by-Step Solution

Answer

The center is (2, 0) and the radius is 4.

Step 1: Understand the Standard Circle Equation

The standard form of a circle's equation is \((x-h)^2 + (y-k)^2 = r^2\), where \((h, k)\) is the center of the circle and \(r\) is the radius.

Step 2: Compare with Standard Equation

Compare the given equation \((x-2)^2 + y^2 = 16\) with the standard form. Here, you can see \(x\) is replaced by \((x-2)\) and \(y\) is simply \(y\), indicating that \(h = 2\) and \(k = 0\).

Step 3: Identify the Radius

The right side of the circle's equation \((x-2)^2 + y^2 = 16\) represents \(r^2\). Therefore, \(r^2 = 16\) and solving for \(r\) gives \(r = \sqrt{16} = 4\).

Step 4: Conclusion of the Center and Radius

Therefore, the center of the circle is \((2, 0)\) and the radius is \(4\).

Key Concepts

Center of the Circle

In order to find the center of any circle from its equation, we need to understand the components of the standard form of a circle's equation, which is formulated as \((x - h)^2 + (y - k)^2 = r^2\). In this equation, \((h, k)\) represents the center of the circle. The values of \(h\) and \(k\) correspond to shifts along the x and y axes, respectively.

To pinpoint the center from a given equation like \((x-2)^2 + y^2 = 16\), you need to compare it with the standard form.

• Note how \((x - 2)\) suggests a horizontal shift by 2 units to the right. This corresponds to \(h = 2\).
• The \(y\) term \((y-0)^2\) can be alternatively written as just \(y^2\), showing no vertical shift. This tells us \(k = 0\).

Thus, the center of the circle in this case is located at the point \((2, 0)\). Understanding the center is crucial as it depicts how the circle is situated on the coordinate plane, helping in graphing and further geometric interpretations.

Radius of the Circle

The radius of a circle is a significant feature that defines the size of the circle. From the standard circle equation \((x-h)^2 + (y-k)^2 = r^2\), \(r\) is the radius. It is derived from the term \(r^2\) on the right side of the equation.

To solve for the radius in our given circle equation \((x-2)^2 + y^2 = 16\), compare it again to the standard form. Here, notice:

• The right-hand side is \(16\), which corresponds to \(r^2\).
• Calculating \(r\) involves taking the square root of \(16\), so \(r = \sqrt{16} = 4\).

This indicates that the circle's radius is 4 units. The radius provides insights into how wide the circle stretches from its center, allowing us to visualize and draw the circle appropriately on a graph. The radius also connects to many practical applications, such as understanding the area and circumference of the circle.

Standard Form of a Circle Equation

Often, when dealing with circle-related problems in algebra, the standard form of a circle's equation is key to unlock all solutions. The equation \((x-h)^2 + (y-k)^2 = r^2\) specifies a circle in a clear and consistent manner.

Here's what each component means:

• \((x-h)^2\) corresponds to how far the circle is shifted horizontally, with \(h\) identifying the x-coordinate of its center.
• \((y-k)^2\) informs about the vertical positioning, where \(k\) is the y-coordinate of the center.
• \(r^2\) is simply the square of the circle's radius, indicating the extent of the circle's size.

When given a circle equation, rewriting or recognizing it in the standard form assists in intuitively determining attributes like the center and radius.

Understanding the standard form equips you with an essential tool in coordinate geometry. It simplifies tackling complex problems, aids in precise graphing, and serves as a foundation for transformations involving circles on the plane.