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Let's work through these problems one by one. The goal is to find the center and radius of each circle and graph its equation. I will also help identify the domain and range for each.

### Problem 124: \( x^2 + y^2 = 1 \)
This is the equation of a circle in standard form: 
\[
(x - h)^2 + (y - k)^2 = r^2
\]
Where \( (h, k) \) is the center and \( r \) is the radius.

- Here, \( h = 0 \), \( k = 0 \) (the center is at the origin), and \( r^2 = 1 \), so \( r = 1 \).
- **Center**: \( (0, 0) \)
- **Radius**: 1

#### Domain and Range:
- **Domain**: The set of all possible x-values. Since the radius is 1 and the circle is centered at the origin, \( x \)-values range from \( -1 \) to \( 1 \). So, the domain is \( [-1, 1] \).
- **Range**: The set of all possible y-values. Similarly, \( y \)-values also range from \( -1 \) to \( 1 \). So, the range is \( [-1, 1] \).

---

### Problem 125: \( (x + 2)^2 + (y - 3)^2 = 9 \)
This is also in the standard form of a circle equation. We can identify:
- \( h = -2 \), \( k = 3 \) (the center is at \( (-2, 3) \)), and \( r^2 = 9 \), so \( r = 3 \).
- **Center**: \( (-2, 3) \)
- **Radius**: 3

#### Domain and Range:
- **Domain**: The center is at \( x = -2 \), and with a radius of 3, the x-values range from \( -2 - 3 = -5 \) to \( -2 + 3 = 1 \). So, the domain is \( [-5, 1] \).
- **Range**: The center is at \( y = 3 \), and with a radius of 3, the y-values range from \( 3 - 3 = 0 \) to \( 3 + 3 = 6 \). So, the range is \( [0, 6] \).

---

### Problem 126: \( x^2 + y^2 - 4x + 2y - 4 = 0 \)
This is not yet in the standard form of a circle equation, so we'll need to complete the square to rewrite it.

First, group the \( x \) and \( y \) terms:
\[
(x^2 - 4x) + (y^2 + 2y) = 4
\]

Now, complete the square for both \( x \) and \( y \):
- For \( x^2 - 4x \), add and subtract \( 4 \) (since \( \left(\frac{-4}{2}\right)^2 = 4 \)).
- For \( y^2 + 2y \), add and subtract \( 1 \) (since \( \left(\frac{2}{2}\right)^2 = 1 \)).

This gives:
\[
(x - 2)^2 - 4 + (y + 1)^2 - 1 = 4
\]

Simplifying:
\[
(x - 2)^2 + (y + 1)^2 = 9
\]

Now the equation is in standard form, so:
- \( h = 2 \), \( k = -1 \) (the center is at \( (2, -1) \)), and \( r^2 = 9 \), so \( r = 3 \).
- **Center**: \( (2, -1) \)
- **Radius**: 3

#### Domain and Range:
- **Domain**: The center is at \( x = 2 \), and with a radius of 3, the x-values range from \( 2 - 3 = -1 \) to \( 2 + 3 = 5 \). So, the domain is \( [-1, 5] \).
- **Range**: The center is at \( y = -1 \), and with a radius of 3, the y-values range from \( -1 - 3 = -4 \) to \( -1 + 3 = 2 \). So, the range is \( [-4, 2] \).

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Would you like further details on any part of this explanation?

Here are some related questions for deeper understanding:
1. How can you determine the equation of a circle if only the center and radius are given?
2. How does completing the square help transform equations into standard circle form?
3. What is the significance of the domain and range for circles in real-world contexts?
4. How do changes in the center affect the graph of the circle?
5. Can the equation of a circle be expressed differently depending on the situation?

**Tip**: When completing the square, remember to balance the equation by adding equal values to both sides!

In Exercises 124-126, give the center and radius of each circle and graph its equation. Use the graph to identify the relation's domain and range.

124. x^2 + y^2 = 1
125. (x + 2)^2 + (y - 3)^2 = 9
126. x^2 + y^2 - 4x + 2y - 4 = 0

Learn how to find the center and radius of circles given their equations and determine the domain and range. This includes completing the square to convert equations into standard form. Suitable for high school students in Grades 9-12.

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