Question

Are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. Write the equation of a circle with radius \(r=\sqrt{5}\) and center (-4,0) in standard from.

Short answer

(x+4)^2 + y^2 = 5

Step-by-step solution

Recall the Standard Form of a Circle's Equation

The standard form of a circle's equation is \( (x-h)^2 + (y-k)^2 = r^2 \) where \( (h, k) \) represents the center of the circle and \( r \) represents the radius.

Identify the Given Information

From the problem, the center of the circle is \( (-4,0) \) and the radius is \( \sqrt{5} \).

Substitute the Values into the Standard Form

Replace \( h \) with \( -4 \), \( k \) with \( 0 \), and \( r \) with \( \sqrt{5} \) in the standard form equation: \[ (x+4)^2 + (y-0)^2 = (\sqrt{5})^2 \]

Simplify the Equation

Simplify \( (y-0)^2 \) to \( y^2 \) and \( (\sqrt{5})^2 \) to \( 5 \). So, the equation becomes: \[ (x+4)^2 + y^2 = 5 \]

Key concepts

Standard Form of a Circle

In algebra, the equation of a circle is often expressed in its standard form. This form makes it easy to see key properties of the circle like its center and radius. The standard form of a circle is:

\[ (x-h)^2 + (y-k)^2 = r^2 \]
Here, \( h \) and \( k \) represent the coordinates of the circle's center, while \( r \) is the radius. By comparing this to other forms, such as the general form, you can directly visualize the primary features of the circle. Here's a breakdown:

• \( (x-h)^2 \): Represents the horizontal distance from the center, \( h \).
• \( (y-k)^2 \): Represents the vertical distance from the center, \( k \).
• \( r^2 \): The squared radius of the circle.

Keeping these points in mind will simplify working with and graphing circles.

Radius of a Circle

The radius is a fundamental part of a circle. It’s simply the distance from the center of the circle to any point on its edge. In the standard form equation \( (x-h)^2 + (y-k)^2 = r^2 \), \( r \) represents the radius.

Given in the exercise, the radius is \( \sqrt{5} \). To use it in the standard form equation, we square the radius:

• \( (\sqrt{5})^2 = 5 \)

Now, the radius squared (\( r^2 \)) is \( 5 \). This makes it easier to substitute into the equation without error. Remember: Always square the radius when fitting it into the standard form.

Center of a Circle

The center of a circle is a crucial concept for its equation. It is represented by the coordinates (\( h \), \( k \)) in the standard form \[ (x-h)^2 + (y-k)^2 = r^2 \]
For interpretation:

• \( h \): The x-coordinate of the center
• \( k \): The y-coordinate of the center

In our exercise, the center is given by (-4, 0). Here is what happens when we substitute these values:

• \( h = -4 \) ➔ (x - (-4)) becomes (x + 4)
• \( k = 0 \) ➔ (y - 0) simplifies to y

Thus, the standard form equation becomes:

\[ (x+4)^2 + y^2 = 5 \]
This finalized equation tells us the center and radius directly for easy graphing.