Problem 3 - Beginning and Intermediate Algebra

Question

Identify the center and radius of each circle and graph. $$(x+2)^{2}+(y-4)^{2}=9$$

Step-by-Step Solution

Answer

The given equation of the circle is \((x+2)^{2}+(y-4)^{2}=9\). Comparing this to the standard circle equation \((x-h)^{2}+(y-k)^{2}=r^{2}\), we find that the center of the circle is at \((-2, 4)\) and the radius is \(3\). To graph the circle, plot the center point and draw a circle with a radius of \(3\) units.

Step 1: Identify the standard form of the given circle equation

The given circle equation is: $$(x+2)^{2}+(y-4)^{2}=9$$Compare this equation to the standard circle equation: $$(x-h)^{2}+(y-k)^{2}=r^{2}$$

Step 2: Find the center (h, k) of the circle

We can see that the given equation has the term \((x+2)^{2}\) which corresponds to \((x-h)^{2}\) in the standard form. Notice that the sign is '+' between x and 2, which means that h is -2, since we have (x-(-2)). Similarly, from the term \((y-4)^{2}\), we find k to be 4. So, the center of the circle is: $$\text{Center } (h, k) = (-2, 4)$$

Step 3: Find the radius r of the circle

Observe that the given circle equation has the term 9 on the right-hand side, which corresponds to \(r^{2}\) in the standard form. The radius r can be found by taking the square root of 9: $$r = \sqrt{9} = 3$$

Step 4: Graph the circle with center and radius

Plot the center (-2, 4) on the graph. From the center, draw a circle with a radius of 3 units. The circle will pass through points 3 units away from the center in all directions.

Key Concepts

Center of a Circle

In the equation of a circle, the center is a crucial point from which every point on the circle is equidistant. The equation of a circle in standard form is \((x-h)^2 + (y-k)^2 = r^2\), where

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

In your example, the equation \((x+2)^2 + (y-4)^2 = 9\) tells us about the center by looking at the parts \((x+2)\) and \((y-4)\).
Translation from the general form reveals that:

• The \((x+2)\) part suggests \(h = -2\) (since it represents \((x - (-2))\)).
• The \((y-4)\) part gives us \(k = 4\).

Therefore, the center of the circle is at the point \((-2, 4)\). The center is the starting point for plotting and understanding the circle's position on a graph.

Radius of a Circle

The radius of a circle defines the distance from its center to any point on its edge. Understanding this concept is simple with the standard equation of a circle: \((x-h)^2 + (y-k)^2 = r^2\). Here,

• \(r\) is the radius, the constant distance from center \((h, k)\) to any point on the circle.

In our particular case, you've observed \((x+2)^2 + (y-4)^2 = 9\).
The right side, 9, equals \(r^2\), so the radius is the square root of 9. Simplifying this gives:

• The radius \(r = \sqrt{9} = 3\).

The radius indicates how far the circle expands around the center point. On a graph, the circle will reach out 3 units from the center, forming a perfect boundary equidistant in every direction.

Standard Form of a Circle Equation

The standard form of the circle equation is a blueprint for identifying key features such as the center and radius. It takes the shape: \((x-h)^2 + (y-k)^2 = r^2\). This representation helps easily pinpoint:

• \((h, k)\), which gives the circle's center.
• \(r\), which gives the circle's radius.

For example, converting any circle equation to this form removes ambiguity and highlights structure, making graphing and analytical work more straightforward. In the exercise given \((x+2)^2 + (y-4)^2 = 9\), compare it with the standard form:

• The terms reveal the center at \((-2, 4)\),
• and the radius as \(3\).

Mastering this form allows the extraction of necessary geometric data and aids in solving more complex mathematical problems related to circles.