Question

Write the standard form of the equation and the general form of the equation of each circle of radius \(r\) and center \((h, k)\). Graph each circle. $$ r=\frac{1}{2} ;(h, k)=\left(\frac{1}{2}, 0\right) $$

Short answer

The standard form is \ (x - \frac{1}{2})^2 + y^2 = \frac{1}{4} \ and the general form is \ x^2 + y^2 - x = 0 \.

Step-by-step solution

- Standard Form of the Circle's Equation

The standard form of a circle's equation with center \( (h, k) \) and radius \( r \) is \[ (x - h)^2 + (y - k)^2 = r^2 \]. Substitute the values \( r = \frac{1}{2} \) and \( (h, k) = \left( \frac{1}{2}, 0 \right) \) into the equation: \[ (x - \frac{1}{2})^2 + (y - 0)^2 = (\frac{1}{2})^2 \].

- Simplify the Standard Form

Now simplify the equation: \[ (x - \frac{1}{2})^2 + y^2 = \frac{1}{4} \]. This is the standard form of the circle's equation.

- General Form of the Circle's Equation

To convert the standard form into the general form, expand the equation: \[ (x - \frac{1}{2})^2 + y^2 = \frac{1}{4} \]. First, expand \( (x - \frac{1}{2})^2 \): \[ x^2 - x + \frac{1}{4} \]. Now add this to \[ y^2 \]: \[ x^2 - x + \frac{1}{4} + y^2 = \frac{1}{4} \]. Subtract \frac{1}{4} from both sides: \[ x^2 + y^2 - x = 0 \].

- Graph the Circle

To graph the circle, plot the center at \( ( \frac{1}{2}, 0 ) \) and draw a circle with radius \frac{1}{2} \ units. Ensure that the circle passes through points that are \frac{1}{2} \ units away from the center in all directions.

Key concepts

Standard Form

The standard form of a circle's equation is very useful because it gives you immediate information about the circle's center and radius. This form is given by
\[ (x - h)^2 + (y - k)^2 = r^2 \]
Here, \( (h, k) \) represents the center of the circle and \( r \) represents the radius. To solve a problem using this form, you simply substitute the given values for the center and radius.

For example, if the radius \( r = \frac{1}{2}\) and the center is at \( \big(\frac{1}{2}, 0\big) \), your equation becomes: \[ (x - \frac{1}{2})^2 + y^2 = \frac{1}{4} \]
This equation tells us exactly where the circle is located and how big it is.

General Form

Once you have the standard form of a circle's equation, you might need to convert it into the general form. The general form looks different and is given by
\[ Ax^2 + Ay^2 + Dx + Ey + F = 0 \]
To convert from standard to general form, you need to expand the equation. For our example:
Start with \[ (x - \frac{1}{2})^2 + y^2 = \frac{1}{4} \]
Expand \( (x - \frac{1}{2})^2 \) to get \[ x^2 - x + \frac{1}{4} \]. Then, add \( y^2 \) and subtract \( \frac{1}{4} \) from both sides:
\[ x^2 - x + y^2 + \frac{1}{4} = \frac{1}{4} \]
\[ x^2 + y^2 - x = 0 \]
This is the general form of the equation.

Radius

The radius \( r \) of a circle is the distance from the center of the circle to any point on the circle. It is an essential part of the circle equation.

In the standard form equation \[ (x - h)^2 + (y - k)^2 = r^2 \]
you can easily see the radius squared on the right-hand side of the equation. For example, if \$r = \frac{1}{2}\text$, you substitute it and get \[ r^2 = (\frac{1}{2})^2 = \frac{1}{4} \].

Knowing the radius helps in sketching and understanding the size of the circle.

Center

The center of the circle is another crucial component and is represented by \( (h, k) \) in the standard form equation. It tells you the exact location of the circle on the coordinate plane.

In our example, the center is at \( \big(\frac{1}{2}, 0\big) \).

This means the circle is shifted \( \frac{1}{2} \) unit to the right from the origin and lies on the x-axis. When plotting the circle, you always start by marking the center point.

Graphing Circles

Graphing a circle requires knowing its center and radius. Once you have the equation (either in standard or general form), follow these steps:

• Plot the center \((h, k)\) on the coordinate plane.
• From the center, measure the radius \( r \) in all directions (up, down, left, and right).
• Draw a curve that smoothly connects these points, forming a circle.

For our example, the center is at \((\frac{1}{2}, 0)\) and the radius is \( \frac{1}{2} \). This means the circle touches the \( y-axis \) at points \((0, 0)\) and \((1, 0)\).

This visualization helps solidify your understanding of the circle's geometry.