Problem 25 - Intermediate Algebra

Question

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

Step-by-Step Solution

Answer

The center is (0, 2) and the radius is 1.

Step 1: Identify the Circle Equation Format

Recognize that the circle's equation is given by the standard form: \((x-h)^2 + (y-k)^2 = r^2\). This equation tells us the center \((h, k)\) and the radius \(r\) of the circle.

Step 2: Extract Center Coordinates

Compare the given equation \(x^2 + (y-2)^2 = 1\) with the standard form \((x-h)^2 + (y-k)^2 = r^2\). Here we see that the term \((x-h)^2\) matches \(x^2\), so \(h = 0\). The term \((y-k)^2\) matches \((y-2)^2\), so \(k = 2\). Therefore, the center of the circle is \((0, 2)\).

Step 3: Determine the Radius

Again, by comparing the given equation \(x^2 + (y-2)^2 = 1\) to the standard form \((x-h)^2 + (y-k)^2 = r^2\), we identify that \(r^2 = 1\). Solving for \(r\), we find \(r = \sqrt{1} = 1\). Thus, the radius of the circle is 1.

Step 4: Graph the Circle

To graph the circle, plot the center at \((0, 2)\) on the coordinate plane. From the center, measure a radius of 1 unit in all directions (up/down/left/right) to sketch the circle. Ensure that the circle remains equidistant (1 unit) from the center at every point along the circumference.

Key Concepts

Center of a Circle

In the world of geometry, the center of a circle is a crucial point. It is defined as the middle point from which all points on the circle's circumference are equidistant. Think of it like the bullseye in a dartboard or the middle of a pizza.
To find the center when given the circle's equation in the standard form

• This standard form formula is \((x - h)^2 + (y - k)^2 = r^2\).
• Here, \(h\) and \(k\) are the coordinates of the circle's center.

In our specific exercise, comparing \(x^2 + (y-2)^2 = 1\) to the standard form reveals that the center is at \((0, 2)\).

• The term \((x-h)^2\) becomes \(x^2\), meaning \(h = 0\).
• Similarly, \((y-k)^2\) becomes \((y-2)^2\), indicating \(k = 2\).

Knowing this allows you to pinpoint the exact middle of your circle on a graph, enhancing your understanding of the circle's position relative to the coordinate plane.

Radius

The radius of a circle is the proverbial key to unlocking its size. This is the distance from the center to any point on the circumference, helping to define the circle's geometry.
Using the equation in standard form, you identify the radius as follows:

• The radius is derived from the equation \((x - h)^2 + (y - k)^2 = r^2\).
• The \(r^2\) in the equation is equal to the value on the other side of the equal sign in the expression.

In our circle's equation, \(x^2 + (y-2)^2 = 1\), we find that \(r^2 = 1\).

• Solving for \(r\), this gives us \(r = \sqrt{1} = 1\).

This means the radius of the circle is \(1\) unit. Knowing the radius is important for sketching the circle or calculating its circumference and area.

Standard Form of a Circle

The standard form of a circle's equation is a convenient toolkit for anyone studying geometry, providing a direct way to understand the core properties of a circle.
When the equation of a circle is presented in the standard form, it appears as:

• \((x - h)^2 + (y - k)^2 = r^2\).
• This formula reveals both the center \((h, k)\) and the radius \(r\) of the circle.

The equation describes a circle centered at \((h, k)\) with a radius \(r\).

• In the exercise, the equation \(x^2 + (y-2)^2 = 1\) matches the standard form, resulting in a center \((0, 2)\) and a radius of \(1\).

This is why converting equations into standard form is beneficial—it simplifies the identification of crucial circle properties that are essential for graphing or solving further mathematical problems.