Problem 72 - Intermediate Algebra

Question

Write each equation in standard form, if it is not already so, and graph it. If the graph is a circle, give the coordinates of its center and its radius. If the graph is a parabola, give the coordinates of its vertex. $$ (x-3)^{2}+(y-2)^{2}=36 $$

Step-by-Step Solution

Answer

The circle has a center at (3, 2) and a radius of 6 units.

Step 1: Identify the Equation Type

The given equation \((x-3)^{2}+(y-2)^{2}=36\) represents a circle because it is written in the form \((x-h)^2 + (y-k)^2 = r^2\).

Step 2: Transform into Standard Form of Circle

The standard form of a circle is \((x-h)^2+(y-k)^2=r^2\), where \((h, k)\) is the center and \(r\) is the radius. The given equation is already in this form: \((x-3)^{2}+(y-2)^{2}=36\).

Step 3: Identify the Center and Radius

Comparing \((x-3)^{2}+(y-2)^{2}=36\) with \((x-h)^{2}+(y-k)^{2}=r^{2}\), we have \(h=3\), \(k=2\), and \(r^2=36\). Thus, the center of the circle is \((3, 2)\) and the radius \(r\) is \(\sqrt{36} = 6\).

Step 4: Graph the Circle

To graph the circle, plot the center at \((3, 2)\) on the coordinate plane. Then, use the radius to draw a circle with a distance of 6 units from the center in all directions.

Key Concepts

Standard Form of a Circle

Understanding the standard form of a circle is crucial for graphing and solving related problems. The standard form equation is expressed as \((x-h)^2 + (y-k)^2 = r^2\). This is the blueprint for all circles in a coordinate plane.

The components of this formula are important:

• \((h, k)\) is the center of the circle.
• \(r^2\) represents the square of the radius.

When given any circle equation, the first step is to check if it's presented in standard form. This helps determine the center and radius easily.

In our example, \((x-3)^2 + (y-2)^2 = 36\) is already in standard form. It's reassuring to have the equation this way because translating it directly provides the critical circle elements such as the center and the radius, essential for graphing.

Finding the Circle Center

The circle center is a key feature needed to visualize and graph a circle properly. To find the center from the standard form equation \((x-h)^2 + (y-k)^2 = r^2\), simply identify the values of \(h\) and \(k\). These coordinates correspond to the center on the coordinate plane.

For example, in the equation \((x-3)^2 + (y-2)^2 = 36\), the values are:

• \(h = 3\)
• \(k = 2\)

Therefore, the center of the circle is at \((3, 2)\).

This point indicates the center from which all points on the circle are equidistant. Always remember, the sign of \(h\) and \(k\) in the equation appears in the opposite sign within the bracket expressions \((x-h)\) and \((y-k)\). This can be a tricky feature to watch out for when identifying the center position from the equation.

Calculating the Circle Radius

The radius is another crucial component of a circle that determines its size. From the circle's standard form equation \((x-h)^2 + (y-k)^2 = r^2\), the \(r^2\) part reveals the radius squared. To find the radius, simply take the square root of this value.

In the equation \((x-3)^2 + (y-2)^2 = 36\), the \(r^2 = 36\).

This means:

• \(r = \sqrt{36}\)
• \(r = 6\)

Thus, the radius of the circle is 6 units.

The radius is the distance from the center to any point on the circle. It is instrumental when plotting the circle since you can measure 6 units in all directions from the center point to draw the complete circle. Remember, all points on the circle are equidistant from its center, and this distance is precisely the radius.