Problem 30 - Algebra A Combined Function

Question

The graph of each equation is a circle. Find the center and the radius, and then graph the circle. See Examples 5 through 7. $$ (x+1)^{2}+(y-2)^{2}=5 $$

Step-by-Step Solution

Answer

Center: (-1, 2), Radius: \(\sqrt{5}\).

Step 1: Identify the Equation Format

Recognize that the equation \((x+1)^{2}+(y-2)^{2}=5\) is in the standard form of a circle's equation: \((x-h)^{2}+(y-k)^{2}=r^{2}\), where \((h, k)\) is the center and \(r\) is the radius.

Step 2: Determine the Center

From the given equation \((x+1)^{2}+(y-2)^{2}=5\), identify the center by comparing it to the standard form. Here, \((h, k)\) is \((-1, 2)\). Therefore, the center of the circle is \((-1, 2)\).

Step 3: Calculate Radius

The value on the right side of the equation \((x+1)^{2}+(y-2)^{2}=5\) represents \(r^{2}\). Therefore, \(r^{2} = 5\), which means the radius \(r = \sqrt{5}\).

Step 4: Graph the Circle

To graph the circle, plot the center at \((-1, 2)\). Then, using the radius \(\sqrt{5}\), draw a circle by moving \(\sqrt{5}\) units in all directions from the center.

Key Concepts

graphing circles

Graphing circles can seem a daunting task at first, but with practice, it becomes much easier. To graph a circle, start by identifying the center and radius from the circle's equation. An equation in the form

• \((x-h)^2 + (y-k)^2 = r^2\),
• where \((h, k)\) is the center and \(r\) is the radius, gives us all the information we need.

The given equation \[(x+1)^2 + (y-2)^2 = 5\]lets us know that the center is at \((-1, 2)\) and the radius is \(\sqrt{5}\).
Once you have these pieces, plot the center on the graph at the point \((-1, 2)\).
Then, from this point, measure a distance of \(\sqrt{5}\) in all directions to draw the circle. Use your compass, or estimate it if you're free-handing. You'll have a perfect circle plotted on your graph plane. Remember to label the x and y axes for clarity.

center of a circle

The center of a circle is a key feature in the equation of a circle. It essentially tells us the exact point from which all the points on the circle are equidistant. In the standard equation form of a circle,
\[(x-h)^2 + (y-k)^2 = r^2\],

• \(h\) and \(k\) are the coordinates of the center:
• This means \((h, k)\) is where the center of the circle lies on the Cartesian plane.

In the equation \((x+1)^2 + (y-2)^2 = 5\),
we compare and find that the center is at \((-1, 2)\).
This is found by substituting \((x+1)^2\) back to \((x-h)^2\) and realizing that \(h = -1\). Similarly for \((y-2)^2\), \(k = 2\). Understanding the center is crucial because it provides a reference point from which you measure the radius to graph the circle.

radius of a circle

The radius of a circle is another fundamental concept, defining the size of the circle. It tells us how far any point on the circle is from its center. In the equation of a circle,

• \(r\) is the radius, described as part of the equation \[(x-h)^2 + (y-k)^2 = r^2\].
• The right side of this equation \(r^2\) gives the square of the radius, which means to find \(r\) itself, simply take the square root.

In our example equation \((x+1)^2 + (y-2)^2 = 5\),
we see \(r^2 = 5\),
indicating that \(r = \sqrt{5}\).
The radius is important for graphing since it is the distance you will use to draw the circle's edge from the center point. Knowing this, you can measure from the center \((-1, 2)\) outwards \(\sqrt{5}\) units in all directions to define the circle's boundary.