Problem 9 - Beginning and Intermediate Algebra

Question

Identify the center and radius of each circle and graph. $$(x-6)^{2}+(y+3)^{2}=16$$

Step-by-Step Solution

Answer

The center of the circle is at the point (6, -3), and the radius is 4 units. To graph the circle, plot the center at (6, -3) and draw a circle with a radius of 4 units.

Step 1: Identify the standard form of circle

The equation provided is already in the standard form of a circle: \((x-a)^2 + (y-b)^2 = r^2\). So, we can directly read the values of a, b, and r.

Step 2: Find the center of the circle

Given that the equation is in standard form, we can see that a = 6 and b = -3. Therefore, the center of the circle is at the point (6, -3).

Step 3: Find the radius of the circle

The given equation is \((x-6)^2 + (y+3)^2 = 16\), which means the radius square r^2 = 16. To find the radius r, take the square root of 16: \[r = \sqrt{16} = 4\]

Step 4: Graphing the circle

To graph the circle, follow these steps: 1. Plot the center of the circle at the point (6, -3). 2. From the center, draw a circle with a radius of 4 units. Now you have successfully graphed the circle with the given equation $$(x-6)^{2}+(y+3)^{2}=16$$. The center of the circle is at (6, -3), and the radius is 4 units.

Key Concepts

Standard Form of a Circle

The standard form of a circle's equation is a structured way of describing a circle in a coordinate plane. This form of the equation is given as \((x-a)^2 + (y-b)^2 = r^2\). Here, \(a\) and \(b\) are the coordinates of the center of the circle, whereas \(r\) represents the radius of the circle.
When we talk about the standard form, we're prioritizing clarity and simplicity. This form makes it easier to directly identify the critical elements of the circle—the center and the radius. Without needing to rearrange or simplify further, students can immediately extract important geometric details from the equation.
Understanding and recognizing this format is crucial when you're tasked with determining a circle's characteristics from its equation. Look for those perfect square terms with \(x\) and \(y\), which point you straight to the location of the circle in the plane.

Center of a Circle

The center of a circle is a critical point that defines its exact position on the coordinate plane. By looking at the standard form equation \((x-a)^2 + (y-b)^2 = r^2\), it's easy to extract the center. The values \(a\) and \(b\) are the coordinates of the circle’s center.
For example, in the equation \((x-6)^2 + (y+3)^2 = 16\), the terms \( (x-6) \, \text{and} \, (y+3) \) directly inform us about the center. Specifically:

• The term \((x-6)\) means the \(x\)-coordinate of the center is 6.
• The term \((y+3)\) suggests the \(y\)-coordinate is -3 (because of the sign change from \(+\) to \(-\)).

Thus, the center of this circle is at the point \((6, -3)\). Knowing the center helps locate the circle and determine areas like peripheral locations (boundary points).

Radius of a Circle

The radius of a circle is the distance from the center to any point on the circle's boundary. In standard form, the radius is obtained from \(r^2\), where \(r^2\) is directly connected to your circle's equation; for example, \(r^2 = 16\).
To find the radius \(r\), simply apply the square root to both sides: \[r = \sqrt{16} = 4\]. This means the radius is 4 units. The radius indicates the circle’s size, allowing you to understand how far the circle extends from the center point in any direction.
A consistent, accurate measurement of the radius is vital not just for graphing but also for deeper mathematical applications. It helps in computing parameters such as circumference and area of the circle, linking geometric intuition and algebraic expressions.